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Student Solutions Manualfor use with
Complex Variables
and ApplicationsSeventh Edition
Selected Solutions to Exercises in Chapters 1-7
by
James Ward BrownProfessor ofMathematics
The University ofMichigan- Dearborn
Ruel V. Churchill
Late Professor ofMathematics
The University ofMichigan
McGrawHill
Higher Education
Boston Burr Ridge, IL Dubuque, IA Madison, Wl New York San Francisco St. Louis
Bangkok Bogota Caracas Kuala Lumpur Lisbon London Madrid Mexico City
Milan Montreal New Delhi Santiago Seoul Singapore Sydney Taipei Toronto
Table ofContents
Chapter 1 1
Chapter 2 22
Chapter 3 35
Chapter 4 53
Chapter 5 75
Chapter 6 94
Chapter 7 118
COMPLEX VARIABLES AND APPLICATIONS" (7/e) by Brown and Churchill
Chapter 1
SECTION 2
1. (a) (V2-/)-/(l-V2i) = V2-/-/-V2=-2i;
(b) (2 -3)(-2,l) = (-4 + 3,6 + 2) = (-1,8);
(0 (3>D(3,-l)(^) = (10,0)(i,
1
L) = (2,l).
2. (a) Re(iz) = Re[i(x + iy)] = Re(-y + ix) = -y = -Im z
;
Im(i'z) = Im[i*(;t + iy)] = Im(-y + a) = x = Re z.
3. (l + z)2 =(l + z)(l + z) = (l+zH + (l + z)z = l-(l + z) + z(l + z)
= l+z+z + z2 = l + 2z + z
2.
4. If z=4±i, then z2 -2z + 2 = (l±i)
2-2(l±t) + 2 = ±2i-2T2i' + 2 = 0.
5. To prove that multiplication is commutative, write
Z1Z2 = (*i.yi)(*2.y2 )= ~ Wa» y^2 + *iy2 )
= (*2*i -yzVp^ H-x^) = (x2,y2 )(xl,yl )= z2zl
.
6. (a) To verify the associative law for addition, write
(Zj + z2 ) + z3 = l(xl ,yl
) + (x2 ,y2 )] + (x3 ,y3
) = (x, + x2 ,v
t+ y2 ) + (*3 ,y3 )
=((*i + x
2 ) + jcj, (y, + y2 ) + y3 )= + (x2 + x
3 ), y, + (y2 + y3 ))
= (^1 » yi ) + (*2 + ^3 . y% + v3) = (*i . yi ) + [(x2 >y* ) + (*3 .^ )]
= z1 +(z2 +z3 ).
2
(b) To verify the distributive law, write
z(zl+ z2 ) = (*,30K*i.yi) + (x2 ,y2 )] = (jc,30(*i + x2 , + y2 )
= (ja, + xx2- yy, - yy2 ,
yjc, + yx2 + xyx+ xy2 )
= (xxl~yy
l+ xx
2 -yy2 ,yx, +xy
l+yx
2+xy
2 )
= ~ yyv yxl+ xy,) + (xx2 - yy2 ,
yx2+ xy2 )
= (x, y)(xl ,yl ) + (x, y)(x2 ,y2 ) = zz
x+zz2 .
10. The problem here is to solve the equation z2 + z + 1 = for z = {x,y) by writing
(x,y)(x,y) + (x,y) + (1,0) = (0,0).
Since
(x2 - y
2 + x + 1, 2xy + y) = (0,0),
it follows that
x2 -y2
+;e + l = and 2xy + y = 0.
By writing the second of these equations as (2x + l)y = 0, we see that either 2x + 1 = or
y = 0. If y = 0, the first equation becomes x2 +x + l = 0, which has no real roots
(according to the quadratic formula). Hence 2x+ 1 = 0, or x = -1/2. In that case, the first
equation reveals that y2 = 3/4, or y = ±V3/2. Thus
SECTION 3
! (a)1 + 2/
|2 - / = (1 + 2Q(3 + 4Q
|
(2 - Q(-5Q _ -5 + lOj -5-10/= 2.
3-4/ 5/ (3-4/)(3 + 4/) (5/)(-5/) 25+
25 5'
> 5/ 5/ _ 5/ _ 1
(1 - 0(2 - 0(3 - ~ (1 ~ 3/)(3 - ~ ^10/ " ~2
5
(c) (l-/)4=[(l-0d-0]
2=(-2/)
2 =-4.
2. (a) (-l)z = -z since z + (-l)z = z[l + (-1)] = z • = 0;
1 1 z z(b) 77- =— - £ = 7 = z fe*0).
1/z z z 1
3
3. (z^Xzj^) = ^[ZaCz^)] = z^iztzjzj = zJC^K)] = z^z^zj] = (z&XZiZj.
6.
Z3Z4
' Z\Z2
1= z.
— £2(z3 *0,z4 * 0).
7.ZyZ _Z2Z \Z2j
VZ\ ^ |=
KZl)
1 = (z2 *0,z*0).c2 /
SECTION 4
w ^=(-73,1), z2 =(V3,o)
4
(c) ^=(-3,1), z2 =(l,4)
Z. + Z,y
(d) z, = + i>i , z2 = *i-%
2. Inequalities (3), Sec. 4, are
Rez<IRezl<lzl and Imz< llmzl< Izl.
These are obvious if we write them as
x<\x\<^x2 + y2
and y<lyl<V*2 +y2
.
3. In order to verify the inequality V2 Izl > IRezI + llmzl, we rewrite it in the following ways:
V2~V*2 +y2
>l*l + lyl,
2(*2 + y
2)>l;cl
2 + 2|jtllyl + lyl2
,
lxl2-2l;tllyl + lyl
2 >0,
(l*l-lyl)2 >0.
This last form of the inequality to be verified is obviously true since the left-hand side is aperfect square.
5
4. (a) Rewrite \z- 1 + il= 1 as |z - (1 - 1)|
= 1. This is the circle centered at 1 - i with radius 1.
It is shown below.
( IX\I \-i J
5. (a) Write I z - 4il+
1
z+ 4/1= 10 as I z - 4il
+
\z - (- 4i)l= 10 to see that this is the locus of all
points z such that the sum of the distances from z to 4i and -4i is a constant. Such a
curve is an ellipse with foci ±4i.
(b) Write \z - \\=\z + i\ as \z - ll=lz - (-i)l to see that this is the locus of all points z such
that the distance from z to 1 is always the same as the distance to -i. The curve is,
then, the perpendicular bisector of the line segment from 1 to -i.
SECTION 5
1. (a) z + 3i' = z + 3i=z-3i;
(b) iz = iz=-iz;
(c) (2 + if = (2 + if = (2 - if = 4 - 4i +
1
2 = 4 - 4/ - 1 = 3 - 4/
;
M I(2z + 5)(V2-0I=I2z + 5IIV2-jI=I2TT5IV2+T = V3I2z + 5I.
2. (a) Rewrite Re(z-/) = 2 as Re[* + i(-y - 1)] = 2, or x = 2. This is the vertical line
through the point z = 2, shown below.
3>
2 x
6
(b) Rewrite I2z - i\= 4 as 2 z --2
radius 2, shown below.
= 4, or = 2. This is the circle centered at — with2
ji/2,
3. Write z, = xl+ iy
land z2 = x2 + ry2 . Then
and
Zr ~z2 = (*, + (y1)-(x2 +%) = (*i-^
2 ) + I'(>'i
-y2 )
= (*i - *2 ) - - ft ) = (*i- Vi)~(x2 - iy
2 ) = zl-z2
ZyZ2 = (xl+ iy
l)(x2 + iy
2 ) = {xxx2- yj2 ) + i(y^
2+ xj
2 )
= (X& -
y
xy2 ) - i{yxx2 + x
ty2 ) = (x1- fy,)^ - iy2 ) = z,z2 .
4. (a) zfaZs = (ziZ2 )z3 = zfo z3 = (*i £2)^3 = *i 22 z3 ;
W z4= z
2z2 = z
2z2 = zzzz = (z z)(z z) = zzzz = z
4.
6. fa)
(b)
ZyZ-2^3 J
z2z3
Z2Z3 Z2 Z3
__lzil_ = Jzi
l_
lz2Z3 l lz2 llz3 l
8. In this problem, we shall use the inequalities (see Sec. 4)
IRezl<lzl and \zx+z2 + z3 |^|£i| + |z2 |
+ M-
Specifically, when lzl< 1,
|Re(2 + z + z3
)|< 12 + z + z
3l < 2+lzl +lz
3l = 2+lzl+lzl
3 < 2 + 1 + 1 = 4.
7
10. First write z4 - 4z
2 + 3 = (z2 - l)(z
2 - 3). Then observe that when Izl=2,
Iz2-ll>|lz
2l-lll| = |lzl
2-l| = l4-ll=3
and
lz2-3l>|lz
2M3l| = |lzl
2-3| = l4-3l=l.
Thus, when lzl=2,
lz4-4z
2+3l=lz
2 -lHz2 -3l>3-l = 3.
Consequently, when z lies on the circle lzl= 2,
1
z4-4z
2 +3 lz4-4z
2+3l 3
1 -<I
11. (a) Prove that z is real <=> z = z.
(<=) Suppose that z = z, so that x - iy = x + iy. This means that i'2y = 0, or y = 0.
Thus z = * + iO = x, or z is real.
(=>) Suppose that z is real, so that z = x + iO. Then z = x - iO = jc + i'O = z.
Prove that z is either real or pure imaginary <=» z2 = z
2.
(^) Suppose that z2 =z2
. Then (jr-i'y)2 =(x + iy)
2, or i4xy = 0. But this can be
only if either x = or y = 0, or possibly jc = y = 0. Thus z is either real or pure
imaginary.
(^) Suppose that z is either real or pure imaginary. If z is real, so that z-x, then
z2 = x
2 = z2
. If z is pure imaginary, so that z = iy, then z2 = (-iy)
2 = (iy)2 = z
2.
12. fa) We shall use mathematical induction to show that
Zi+z2 +-"+z„ =z1+z2 +---+zn (n = 2,3,...).
This is known when n = 2 (Sec. 5). Assuming now that it is true when n = m, we maywrite
Zt+Z2 +*"+Zm +Zm+1 - (Z
t+Z2 + -,-+£m ) + Zm+l
= (z1+z2 +-+zm ) + zm+1
= Zt+ Z2 H l"Zm ) + Zm+1
= Zj + Z2 H l"Zm + Zm+1 .
(b) In the same way, we can show that
^z1 ---zn =zlz1 --zn (n = 2,3,...).
This is true when n = 2 (Sec. 5). Assuming that it is true when n = m, we write
^lZ2' '
'£mZm+l ~ (Z1Z2 "'Zm )Zm+l ~ (^2 "
'Zm )
= (^1^2"
'2m )^m+l
= ^2 '*
' ^nAr
14. The identities (Sec. 5) zz =\z\2and Rez = -^-^ enable us to write lz-z l= R as
2
(z-Z )(z-Z ) = i?2
,
zz-(z5, + z!Q+Zo5> =/?2'
lzl2-2Re(zz ) + lz l
2 = /?2
.
15. Since x - -^-^ and y = the hyperbola x2 -y2 = 1 can be written in the following
ways:
. _n2 / _\2z+z\ fz-z
= 1,
2 ; 1 2i
z2 + 2zz + z
2. z
2 -2zz+z 2
= 1,
2z2 + 2z
2_
z2+z
2=2.
SECTION 7
1. (a) Since
argl _2 -2i '
= 3X8
1
~ afg(~2 ~ 2/) '
one value of argj*' —
jis
y ~f~~£~J'or Consequently, the principal value is
9
(b) Since
arg(V3-i)6=6arg(V3-0,
one value of arg(V3 -i*)6
is 6f™\ or -it. So the principal value is -it + 2n, or it.
4. The solution d = n of the equation \e'e -11= 2 in the interval < < In is geometrically
evident if we recall that e'e
lies on the circle \z\= 1 and that le'* - II is the distance between
the points eieand 1. See the figure below.
5. We know from de Moivre's formula that
(cos + isin 0)3 = cos30 + isin30,
or
That is,
cos3 + 3cos
20(/'sin 0) + 3cos 0(isin 0)
2 + (/sin 0)3 = cos30 + isin 30.
(cos3 - 3cos 0sin
z0) + i'(3cos
z0sin - sin
30) = cos30 + isin30.
By equating real parts and then imaginary parts here, we arrive at the desired trigonometric
identities:
(a) cos30 = cos3 0-3cos0sin2
0; (b) sin30 = 3cos20sin0-sin
30.
8. Here z = re'9
is any nonzero complex number and n a negative integer (n = -l,-2,...).
Also, m = -n = l,2 By writing
and
(z-l
)
m
=[l^~6)
J =[-JeK-mB) =\e i(-~me
\
we see that (zm
) =(z )
m. Thus the definition z" = {z )
mcan also be written as
z"=(zmT
l
-
10
9. First of all, given two nonzero complex numbers zi and z 2 , suppose that there are complexnumbers c
xand c 2 such that Zj = qc
2and z^ = c
tc2 . Since
IzJHqllcJ and IzjIHcJIqIMcJIql,
it follows that IzJ^ZjI.
Suppose, on the other hand, that we know only that 1^1=^1. We may write
Zi = r{exp(i6
l ) and z2 = r, exp(/02 ).
If we introduce the numbers
q = rxexpl i
2
Jand c2 = expl i
*
we find that
2 J'
c,c2 = rt exp^i-^-Jexpl
J= r
texp(i^) = z,
and
i+ilvi ,.0i-0qc2 = rtexp| i
-^y*Jexp^-i
|= n exp 2
= z2 .
That is,
Zi = qc2 and z2 = c,c2 .
10. If 5, = l + z+z2
+"-+z", then
5 - zS = (1 + z + z2+• • .+z" ) -
(Z + z2 + z
3 +. . . ) = i _ £"+1
l-z"+1
Hence S =— .provided z*l. That is,
1-z
1 — 7"+1
H-z + z2+-+z"=— (z*l).
1-z
Putting z = e'e(0 < < 2tt) in this identity, we have
1 J(n+l)6
l + em +ei20+-+e ine =
l ~ e
\-e ie
Now the real part of the left-hand side here is evidendy
l + cos0 + cos20+-+cosn0;
and, to find the real part of the right-hand side, we write that side in the form
11
l-exp|7(n + l)0]
l-exp(/0)
exp
exp(-!)
exp^-if)
-exp\(2n + l)0"i- —
2
expH) -exp®which becomes
9 . . (2n + l)0 . . (2n + l)0cos— i sin— cos ism •
2 2 2 2 1
-2i'sin—2
or
r . , (2n + l)0l J (2n + l)0lI sm— + sin -
—
^ j+ 1 1 cos— - cos -
—
^ j
2sin—
The real part of this is clearly
. (2/i + l)0
i +sin
-ir-2 2sin^
2
and we arrive at Lagrange's trigonometric identity:
(2n + l)0
1sm
1 + cos + cos20+- • -+cosn0 = — + ^
—
2 2sin—
(0< 0<2;r).
12
SECTION 9
1. (a) Since 2/ = 2exp i\^ + 2kn (k = 0,±1,±2,...), the desired roots are
(2/)1/2=V2exp
That is,
and
= i+i
c being the principal root These are sketched below.
(* = <U).
(b) Observe that 1-V3i = 2exp i\ —j + 2Jbr (£ = 0,±1,±2,...). Hence
(l-V3/)1/2=V2exp
The principal root is
i'I -^ + kn
c = V2V''*'6 = V2[ cos| - /sin| ] = V2
,2 2 J
V3-i
V2'
and the other root is
V2
These roots are shown below.
(A: = 0,1).
13
2. (a) Since -16 = 16exp[i(;r + 2*7r)] (k = 0,±1,±2,...), the needed roots are
(-16)1/4 = 2exp ,
it kiti — +—4 2
(* = 0,1,2,3).
The principal root is
c =2,- = 2(cosf + / Si„l] = 2(-^ + -^j = V2(l +i).
The other three roots are
and
cx= {ij^y*11 = c
Qi = V2 (i + /)/ = -V2 (i - o,
c2 =(2O*'* = -c =-V2(l + i),
c3= (2e'*
/4)e»*
12 = c (-i) = V2 (1 + OH) = V2(l - /).
The four roots are shown below.
(b) First write -8 -8V3/ = 16exp ^-~ + 2kn^ (* = 0,+l,±2,...). Then
(-8-8V3i)l/4 =2exp ,
it kit(* = 0,1,2,3).
The principal root is
c =2e-,Vr/6 =2 cos^-/sin- =2
I 2 2= V3-i.
The others are
c2 =(2,-/6)e
i'r
= -c =-(V3-0,
c3 = (2e-'wy 3*'2 = c (-/) = -(1 + V3/).
These roots are all shown below.
(a) By writing -1 = lexp[i(?r + 2kn)] (k = 0,±1,±2,. . .), we see that
(-1)1/3 = exp
The principal root is
., % 2kiti — +
3 3
Jt . . jc 1 + V3ic =e = cos— + jsin— =
The other two roots are
and
c, = e'* = -l
„ _ jsxii _ nx-ixn „ K . . 7t 1-V3ic2= e =e e = cos isin— = ,
All three roots are shown below.
15
(b) Since 8 = 8exp[i(0 + 2for)] (k = 0,±1,±2,...), the desired roots of 8 are
81/6 =V2exp(/^ (* = 0,1,2,3,4,5),
the principal one being
The others are
c =V2.
= V2V- = V2(cos| + /sin|] = Vlfi +^/V^.v 2 2 ,
: (V^"*'3)** = V^cosy - isin-0-1) = -V2
1 V3
2"T
V2
N1-V3/
V2*
c3 =j2e'*=-j2,
c4 =(j2eM3
)eb, = -c
i
=-
and
1 + V3/
V2'
c5 =(V2V
2*V* = -c2=i^.
V2
All six roots are shown below.
The three cube roots of the number z = -A^2 + A-Jli = 8exp^/-^j are evidently
(zor=2exp[I + 2|E)] (* = 0,U).
In particular,
= 2exp^-|j = V2(l + i).
16
1 H" "^3/With the aid of the number o3
=, we obtain the other two roots:
v 2 J V2
c2 =c ©3 =(c Q)3)fi)
3= •(V3 + 1) + (V3-1)/
V2
-1 + VIA _(V3-1)- (V3 + l)i
2 J V2
5. (a) Let a denote any fixed real number. In order to find the two square roots of a + i in
exponential form, we write
Since
we see that
A = \a + i\ =Var+T and a = Arg(a + i).
a + i = Aexp[i(a + 2kn)]
(a + i)V2 =VAexp
That is, the desired square roots are
ify + Jbr
VaV "2and VaV^V* = -4Aeia,
\
(* = 0,±1,±2,...),
(* = 0,1).
cc it
(b) Since a + i lies above the real axis, we know that < a < n. Thus <— <— , and this
2 2
tells us that cosf ^ ] > and sinf )> 0. Since cosa =— , it follows that
\2 \2 A
a /1 + cosor 1 [. ~a ^fA + a
and
. asin
2 V 2 -7211 A 42-fA'
1 - cos a 1 |^__a _ VA-a
Consequently,
±VAVa/2 =±VA^cos| + isin|j = ±VA^
= ±-^(VA + a+iVA-a).
VA + a . VA-fl ^|
^Va +iV2VA J
17
6. The four roots of the equation z4 + 4 = are the four fourth roots of the number -4. To
find those roots, we write -4 = 4expD'(;r + 2ibr)] (& = 0,±1,±2,...). Then
(-4)1/4 =V2exp
To be specific,
(f + Y)l = V2^/4e^2(* = 0.1,2,3).
c = V2V™ = V2fcos^ + /sin^l = V2 f-L + i
*
V 4 4j VV2 V2j
q =c e"c/2
=(l + i)i = -l + i,
c2 =c e"r
= (l + 0(-l) = -l-i,
= l + i,
c3 = Coei3*/2 = (1 + /)(-/) = 1-/.
This enables us to write
z4 +4 = (z-
c
)(z - q)(z - c2 )(z - c3 )
= [(z - q)(z - c2 )] • [(z - c )(z - c3 )]
= Kz + 1) - i][(z + 1) + 1] • [(z - 1) - i][(z - 1) + 1]
= [(z + l)2+ l][(z-l)
2+ l]
= (z2 + 2z + 2)(z
2-2z + 2).
7. Let c be any nth root of unity other than unity itself. With the aid of the identity (seeExercise 10, Sec. 7),
.2 , ,l-Z"
we find that
l+z + zz+-+z"-
I =-^- (z*l),l-z
l + c + c2+...+c«-i = l_£l =: J__L =
1-c 1-c
9. Observe first that
(z-r^^exp^^) 1 i(-0
WeXP_-2tor) 1 i(-0) i(-2kn)= -7=exp -exp- -
m "ilr m m
18
and
,.-wm_Ji i(-e + 2kn) 1 i(-8) i(2kn)(z ) -T-exp = -
7=rexp-—-exp- -,
where Jfe = 0, 1, 2, . . . ,m - L Since the set
i{-2kn)exp-i L
(* = 0,l,2,...,ro-l)m
is the same as the set
i{2kn)exp^ L
(* = 0,l,2,...,m-l),m
but in reverse order, we find that (z1/m
)1 = (z
_1
)1/m
.
SECTION 10
1. (a) Write I z - 2 + il£ 1 as I z - (2 - i)l£ 1 to see that this is the set of points inside and on the
circle centered at the point 2-i with radius 1. It is not a domain.
O
(b) Write I2z + 3I>4 as > 2 to see that the set in question consists of all points
exterior to the circle with center at -3/2 and radius 2. It is a domain.
19
(c) Write Imz > 1 as y > 1 to see that this is the half plane consisting of all points lying
above the horizontal line y = 1. It is a domain.
y
y =i
X
(d) The set Imz = 1 is simply the horizontal line y = 1. It is not a domain.
y
y = \
X
(e) The set < argz < — (z * 0) is indicated below. It is not a domain.4
(f)The set Iz - 4l£l zl can be written in the form (x - 4)
2 + y2 1 x
2 + y\ which reduces to
*^2. This set, which is indicated below, is not a domain. The set is also
geometrically evident since it consists of all points z such that the distance between z
and 4 is greater than or equal to the distance between z and the origin.
20
4. (a) The closure of the set -it < argz < n (z # 0) is the entire plane.
1
(b) We first write the set IRezklzl as \x\<^Jx2 +y2
, or x2 <x2 + y
2. But this last
inequality is the same as y2> 0, or \y\> 0. Hence the closure of the set IRezklzl is the
entire plane.
(c) Since - =— = 7-7 = -5
—
-rr, the set Re - < - can be written as —. = < -, orz zz \z\
2 x2 +y2VzJ 2 x
2 +y22
(x2 -2x) + y
2>0. Finally, by completing the square, we arrive at the inequality
(x - 1)2 + y
2 > l2
, which describes the circle, together with its exterior, that is centered
at z = 1 with radius 1. The closure of this set is itself.
21
(d) Since z2 = (x + iyf =x2 -y2
+ ilxy, the set Re(z2) > can be written as y
2 < x2
, or
\y\<\x\. The closure of this set consists of the lines y = ±x together with the shaded
region shown below.
Since every polygonal line joining ztand z2 must contain at least one point that is not in S, it
is clear that S is not connected.
8. We are given that a set S contains each of its accumulation points. The problem here is to
show that S must be closed. We do this by contradiction. We let Zq be a boundary point of
S and suppose that it is not a point in S. The fact that z is a boundary point means that
every neighborhood of zQ contains at least one point in S; and, since z is not in S, we see
that every deleted neighborhood of S must contain at least one point in S. Thus Zq is an
accumulation point of S, and it follows that z is a point in S. But this contradicts the fact
that Zq is not in S. We may conclude, then, that each boundary point Zq must be in S. That
is, S is closed.
5. The set S consists of all points z such that lzl< 1 or lz - 2I< 1, as shown below.
22
Chapter 2
SECTION 11
1. (a) The function f(z) = 2
*is defined everywhere in the finite plane except at the
z +1
points z = ±i, where z2 + 1 = 0.
(fc) The function /(z) = Arg( -]is defined throughout the entire finite plane except for the
point z = 0.
z(c) The function /(z) = is defined everywhere in the finite plane except for the
z + z
imaginary axis. This is because the equation z + z = is the same as x = 0.
(d) The function /(z) = -— is defined everywhere in the finite plane except on the1 — Izl
circle Izl = 1, where 1 -Izl2 = 0.
3. Using x = and y = ^—-, write
f(z) = x*-y2 -2y+ i(2x - 2xy)
. fa±tf + ftdg, + _ + + _ fet Sfe -o
_2 -2 2 -2
= ir + 4r + 2iz-^- + ^- = z
2+2/z.
2 2 2 2
SECTION 17
5. Consider the function
/w-if l
-'x + iyY (z*0),
where z = x + iy. Observe that if z = (jc,0), then
x + iON2
and if z = (0,y),
23
But if z = (*,*),
/fe) = l,I^ =137.1
This shows that /(z) has value 1 at all nonzero points on the real and imaginary axes but
value -1 at all nonzero points on the line y = x. Thus the limit of /(z) as z tends to
cannot exist.
4z2
10. (a) To show that lim -r = 4, we use statement (2), Sec. 16, and write(z - 1)
4 -P2
lim ,kZ
\-, = lim——-? = 4.
V-»i (z-l)J(b) To establish the limit lim ——
—
j = <», we refer to statement (1), Sec. 16, and write
lira -7—^—r = lim (z - 1)3 = 0.
*-»il/(z-l)3
«-»
z2 + l
(c) To verify that lim = °°, we apply statement (3), Sec. 16, and writez-1
i-ilim —^-j = lim =- = 0.
«^fly j^» l+z
2
11. In this problem, we consider the function
7-(z ) = f£±A (ad- be *0).CZ + fi?
fa) Suppose that c = 0. Statement (3), Sec. 16, tells us that lim 7(z) = °° since
lim = hm = — = 0.«-»° T(l/z) »-»o a + bz a
24
(b) Suppose that c * 0. Statement (2), Sec. 16, reveals that lim T(z) = — since*-»- c
limT|'il = lima+ fe a
\zJ c + dz c
Also, we know from statement (1), Sec. 16, that lim T(z) = 00 sincel-*-d/c
Um -i-= lim £+4,0.7(2) z-»-rf/c az + &
SECTION 19
1. (a^ If /(z) = 3z2 -2z + 4,then
/'(z) = ^(3z2-2z + 4) = 3^-z
2 -2-^-2 + 4-4 = 3(2z) - 2(1) + = 6z - 2.
fife az az az
If /(z) = (l-4z2)
3,then
/'(z) = 3(1 - 4zT T-d - 4* > = 3(! " 4z2f(-8z) = -24z(l - 4z
2)
dz
rL)(2z + l)f(z-D-(z-l)f(2z + l)
(2, + 1)(1). (z _ 1)2 _ 3
JK)(2z + l)
2(2z + l)
2(2z + l)
2
(d) If /(z) =(1+
f)4
(z^0),thenz
^An^^_ a + z2yd 2
dz
(z2
)
2(z
2
)
2nz) = ' ^ <1 + Z )~ (l + Z ) g = z
24(lH-z
2)
3(2z)-(lH-z
2)42z
_ 2z(l + z2)
3[4z
2-(l + z
2)] _ 2(l + z
2
)
3(3z
2 - 1)
25
3. If /(z) = l/z (z*0), then
1 1 -AzAw = f(z + Az)-f(z) =z + Az z (z + Az)z
Hence
/'(z) = lim— = lim — =—\r.A*->o *z-*0(z + Az)z z
We are given that f(z ) = g(z ) = and that f\z ) and g'(z ) exist, where g'(z )*0.According to the definition of derivative,
Similarly,
Thus
rfe) = lira /Mzfe) = lim/w.
«-*«o Z - Zq Z-Zq
g-(z ) = lim^)-^ ) = lim-^-.
lim = UrnfWb-**) -_\^Z)I{Z-^
_ f(Zo)
*-*zog(z) s(z)/(z-z ) lim^(z)/(z-z ) ^'(z,,)'
SECTION 22
1. (a) /(z) = z= x — iy. So u = x, v = -y.
Inasmuch as = Vj, => 1 = -1, the Cauchy-Riemann equations are not satisfied
anywhere.
(b) f(z) = z-z=(x + iy)-(x-iy) = + i2y. So u = 0, v = 2y.
Since mx = vy=> = 2, the Cauchy-Riemann equations are not satisfied anywhere.
(c) f(z ) = 2x + ixy2
. Here u = 2x, v = xy2
.
ux = vy=> 2 = 2xy => xy = 1-
w = -vT=> = -y2
=> y = 0.
Substituting y = into Ay = 1, we have = 1. Thus the Cauchy-Riemann equations donot hold anywhere.
(d) /(z) = exe~'
y - e*(cosy - isiny) = excosy - ie
xsiny. So u = e
xcosy, v = -e
xsiny.
«x = vy=> cosy = -e
xcosy => 2e* cosy = => cosy = 0. Thus
y =^ + nn (n = 0,±l,±2,...).
uy= -v* =* -e'siny = e'siny => 2e*siny = => siny = 0. Hence
y-nit (« = 0,±1,±2,...).
Since these are two different sets of values of y, the Cauchy-Riemann equations cannotbe satisfied anywhere.
26
3. (aj /(z) = - = -.- =— =— _ + Soz z z Izl x +y2
x2 +y2
Since
x a -yu = —: t and v = , , .
x2 + y
2x
2 + y2
/'(z) exists when z 0. Moreover, when z * 0,
y2 -*2
. 2xy
(*2 +v2
)
2+V+/)2 "(x
2 +y2)
2
» , . y2 -*2
,. 2xy x
2-i'2jcy-y
2
_ (*-») = (gr _ ay \_
(x2 +y2
)2
(zz)2
(z)2(z)
2z2-
(fc) /(z) = jc2 + /y
2. Hence u = x
2and v = y
2. Now
ux = vy=> 2x = 2y => y = jc and M
y= -v, => = 0.
So /'(z) exists only when y = x, and we find that
f\x + ix) = ux (x,x) + ivx (x,x) = 2x + 1O = 2x.
W f(z) = z Im z = (x + iy)y = xy + iy2
. Here u = xy and v = y2
. We observe that
wx = vy=> y = 2y => y = and «
y= -vx * = 0.
Hence /'(z) exists only when z = 0. In fact,
/'(0) = Mje (0, 0) + iv, (0, 0) = + iO = 0.
4. (flj /(z) = -^ = ^cos4^ +^~sin40j (z*0). Since
4 4ru
r=— cos40 = vg and ue =—-sin40 = -rv
r
/is analytic in its domain of definition. Furthermore,
f'(z) = e-w(u
r+ iv
r )= e-
w^~cos40 + i^-sin4aj
= ~e- ie(cos46-isin4d) = --^"'V'48
r r-4 4 4
r3e,5B
(re,e
f
(b) f(z) = -Jre''2 = Vr cos— + n/r sin— (r > 0, a < < a + In). Since
2* 2*
^ , .
rur=—cos- = ve and ue =-—sm- = -rv
r ,
/is analytic in its domain of definition. Moreover,
f\z) = e-w(u
r + ivr )=MrT--cos| + irVsiiif
-— «"wfcos- + /sin-) = A=e-Wewn2^ I 2 2 J 2V^
1= 1
2^eien ~2f(z)'
(c) f(z) = e~ecos(lnr) + ie-
esin(lnr) (r>0,0< 0<2tt). Since
V „ ' Vy
1
U V
rur=-e~e
sin(ln r) = ve and ug =-e~ecos(ln r) = -rv
r
/is analytic in its domain of definition. Also,
f'(z) = e-ie(ur +ivr )
= e-w e~ sin(lnr)
|^.
gacos(lnr)
r r
= -^fe~ecos(ln r) + ie~
esin(lnr)l =
re L J z
When /(z) = *3 + i(l - yf, we have k = x 3
and v = (1 - y)\ Observe that
ux = vy=> 3x
2 = -3(1-yf=*x2+ (1-yf=0 and w
y=-
Vje =>0
Evidently, then, the Cauchy-Riemann equations are satisfied only when x = and y = 1.
That is, they hold only when z = i. Hence the expression
f'(z) = ux + ivx = 3x2 + iO = 3x
2
is valid only when z = i, in which case we see that /'(/) = 0.
Here u and v denote the real and imaginary components of the function / defined by meansof the equations
Az) =t— when z*0,z
when z = 0.
Now
xl +y2x
2 + y2
when z * 0, and the following calculations show that
",(0,0) = v, (0,0) and «y (0,0)
= -v,(0,0):
Ax-»0 AjC A*->0 AjC
w(0 + Ajc,0)--«(0,0)
Ax
w(0,0 + Ay)- «(0,0)
Ay
v(0 + Ajc,0)- v(0,0) _Ax
v(0,0 + Ay)- v(0,0)
Ay-»0 Ay Ay-»0 Ay
A"->0 Ax
Ay-»0 Ay Ay->0 £y
Equations (2), Sec. 22, are
ux cos + uysin
-w,rsin0 + iycos0
= "r>
Solving these simultaneous linear equations for ux and uy
, we find that
a sin 6 , . n cos0ux = u
rcosO-ug and u
y=ur
smd + ue
Likewise,
r
a sin0 cos0= v
rcos -
v
e and vy= v
rsm0 + vg .
Assume now that the Cauchy-Riemann equations in polar form,
rur=vg ,
ue =-rvr ,
are satisfied at z„. It follows that
a sin 6 cos 6,
. _ . - cos0wx = i/
rcos0-we
= ve + vsin0 = vr smd+ vg = vv ,
r r ry
. a ,cos0 sin0 _ f . sin0\
My= «
rsin d + ue
= ve vrcos0 = -l v
rcos0-ve \
= ~vx-
(a) Write f(z) = u(r, 0) + iv(r, 0) . Then recall the polar form
r«r=ve ,
M„=-rvr
of the Cauchy-Riemann equations, which enables us to rewrite the expression (Sec. 22)
f'(z ) = e-ie(ur +ivr )
for the derivative of/at a point z = (r , O ) in the following way:
f'(Zo ) = e~i0(-vg --u(
A =—lg (ue + ivg ) = —(ue + ivg ).
\r r J re zn
30
(b) Consider now the function
With
f{^\ _ 1 _ 1 _ 1 -is _ 1 /„ „ Q . • Q\ cos sin 6j\Z)-- =—ig=-e = -(cos 6 -ism 6) = 1.
z re r r r r
,.(*. Q\ C0S & At as S"VS\Bu{r, d) = and v(r, 0) =
,
r r
the final expression for /'(z ) in part (a) tells us that
£K \~
l ( sin # .cos0/ (z) =— 1
z \ r r
1fcos - /sin 6
z\ r
\ r J zKre'l z2
when z*0.
10. (a) We consider a function F(x,y), where
z + zX = and y =
z-z2 '
2i
Formal application of the chain rule for multivariable functions yields
dz dxdz dy dz dx \2 J dy \ 2i)~ 2{dx+l
dy
(b) Now define the operator
dz 2\dx dy j
suggested by part (a), and formally apply it to a function f(z) = u(x,y) + iv(x,y):
dz 2{dx dy) 2dx 2dy
=fa +*J +fa + ivy)= \[{ux - v
y ) + i(vx + uy )].
If the Cauchy-Riemann equations uz = vv , u = -vr are satisfied, this tells us thaty y *
df/dz = 0.
31
SECTION 24
1. (a) f(z) = 3x + y + i(3y - x) is entire since
U V
ux =3 = vy
and uy=l = -vx .
(b) f(z) = sinx cosh y + icosx sinhy is entire since> „ ' v
v,
U V
ux = cosx cosh y = vy
and uy= sinxsinhy = -vx .
(c) f(z) = e~y sinx - ie~
y cosx = e~y s'mx + i(-e~y cosx) is entire since
' v ' > , ,
V
ux = e~y cosx - vy
and uy= -e~y sinx = -vx .
(d) f(z) = (z2 - 2)e'
xe"y is entire since it is the product of the entire functions
g(z) = z2 -2 and h(z) = e~
xe~
ty = e~x(cosy - isiny) = e~
xcosy + i(-e~
xsiny).
' v ' > v'
U V
The function g is entire since it is a polynomial, and h is entire since
ux = -e~xcosy = v
yand u
y= -e~
xsiny = -vx .
2. (a) /(z) = xy + iy is nowhere analytic since
U V
ux = vy=>y = l and u
y=-vx =>x = 0,
which means that the Cauchy-Riemann equations hold only at the point z = (0,1) = i.
(c) f(z) = eye
lx = ey (cosx + /sin x) = gycosj: + iVsinjc is nowhere analytic since
U V
Mx = vy=* -ey sinx = ey sinx 2ey sinx = => sinx =
and
= -v, => ey cosx = -ey cosx => 2^ cosx = => cosx = 0.
More precisely, the roots of the equation sinx = are nit (n = 0,±l,±2,...), and
cos /iff = (-1)" * 0. Consequently, the Cauchy-Riemann equations are not satisfied
anywhere.
7. (a) Suppose that a function f(z) = u(x,y) + iv(x,y) is analytic and real-valued in a domain
D. Since f(z) is real-valued, it has the form f(z) = u(x,y) + i0. The Cauchy-Riemann
equations ux =vy,u
y= -vx thus become ux =0,uy
= 0; and this means that u{x,y) = a,
where a is a (real) constant. (See the proof of the theorem in Sec. 23.) Evidently, then,
f(z) = a. That is, / is constant in D.
32
(b) Suppose that a function / is analytic in a domain D and that its modulus is
constant there. Write |/(z)| = c, where c is a (real) constant. If c = 0, we see that
f(z) - throughout D. If, on the other hand, c * 0, write f(z)7(z) = c2
, or
/(*) = •
C"
mSince f(z) is analytic and never zero in D, the conjugate f(z) must be analytic in D.
Example 3 in Sec. 24 then tells us that f(z) must be constant in D.
SECTION 25 (Yl
1. (a) It is straightforward to show that uxx +uyy=0 when u(x,y) = 2x(l-y). To find a
harmonic conjugate v(x,y), we start with ux (x,y) = 2-2y. Now
ux =vy=>v
y=2-2y=> v(x,y) = 2y-y2 + </>(x).
Then
uy= -vx => -2x- -<p'(x) => 0'(jc) = 2* = *
2 + c.
Consequently,
v(;c,y) = 2y-y2+ (jc
2 + c) = x2 - y2 + 2y + c.
(b) It is straightforward to show that uxx + uyy=0 when u(x,y) = 2x - x 3 + 3xy
2. To find a
harmonic conjugate v(x,y), we start with w^Oc.y) = 2-3*2 + 3y2
. Now
ux = Vy ^>vy=2-3x2
+ 3y2=» v(*,y) = 2y- 3x
2y + y
3 + <j)(x).
Then
uy= -vx => 6xy = 6xy - <p'(x) => <p'(x) = 0=> 0O) = c.
Consequendy,
v(jc,y) = 2y - 3x2y + y
3 + c.
fcj It is straightforward to show that uxx +uyy=0 when «(jc,y) = sinhjcsiny. To find a
harmonic conjugate v(x,y), we start with ux (x,y) = coshx sin y. Now
= vy
vy= coshxsiny => v(x,y) = -coshxcosy + <p(x).
Then
«j, = -vx => sinhxcosy = sinhxcosy - <j)'(x) => 0'(jc) = => 0(*) = c.
Consequently,
v(*,y) = - coshxcosy + c.
(d) It is straightforward to show that m„+w=0 when u{x,y) = , - , . To find a
harmonic conjugate v(x,y), we start with ux (x,y) = r~^~TT- Now(x
2 + y2)2
Then
= v, => vy= ~
(jc2
2
+ 2)2=> v(*,y) = -^£-5.+ ^(jc).
2 2 2 2
"y= "Vx ^ (x
2+y
2
)
2=
(x2 + y
2)
2 ~ ^ *'(jc) = 0=*
=°-
Consequendy,
v(x,y) =— T + c.
x +y
Suppose that v and V are harmonic conjugates of u in a domain D. This means that
ux = vy ,
uy= -vx and ux = Vy ,
uy= -Vx .
If w = v - V, then,
wx =vx -Vx =-uy+Uy=0 and w
y=v
y-V
y=ux -ux
= 0.
Hence w(jc,y) = c, where c is a (real) constant (compare the proof of the theorem in Sec.
23). That is, v(x,y)-V(x,y) = c.
Suppose that u and v are harmonic conjugates of each other in a domain D. Then
ux = vy ,
Uy = -vx and vx = uy ,
vy= -ux .
It follows readily from these equations that
ux — 0, uy= and vx = 0, v
y= 0.
Consequently, w(*,;y) and v(x,y) must be constant throughout/) (compare the proof of the
theorem in Sec. 23).
The Cauchy-Riemann equations in polar coordinates are
rur=ve and ue - -rvT
.
Now
rur= ve => run + u
r= v
er
and
Thus
run + rur+ ugg =rv6r
- rvr6 ;
and, since v^. = vr6 , we have
r2un + rur + Ugg = 0,
which is the polar form of Laplace's equation. To show that v satisfies the same equation,
we observe that
1 1 1
"e = -rvr=> v
r=— ug
=> v„ =—ug—
K
frr r r
and
rur=vg =*vgg
= rur0 .
Since uBr= urg , then,
r2vn + rvr + vee = ue - ru^ -ue + rur8 = 0.
If w(r,0) = lnr, then
rV +mr + ugg = r2
[
~J
+ rfi 1 + = 0.
This tells us that the function w = In r is harmonic in the domain r>O,O<0<27T. Now it
follows from the Cauchy-Riemann equation rur= vg and the derivative ur~~ mat ve ~ ^
thus v(r,6) = + <j>(r), where 0(r) is at present an arbitrary differentiable function of r.
The other Cauchy-Riemann equation ug =-rvrthen becomes O = -r0'(r). That is,
<j>\r) = 0; and we see that <j>(r) = c, where c is an arbitrary (real) constant. Hence
v(r, 6) = 8 + c is a harmonic conjugate of u(r, 6) = In r.
35
Chapter 3
SECTION 28
1. (a) exj>(2±3m) = e2ex$(±3m) = -e
2, since exp(±3;ri) = -l.
2 + m ( IV m\ r( n . . n\(b) exp—^— = 1 exp- II exp— 1 = V«l cos- + isin- I
(c,) exp(z + 7ri) = (expz)(exp7ri) = -expz, since exp7ri = -l.
3. First write
exp(z ) = exp(* - iy) = exe~
!y = excos y - ie
xsin y,
where z = x + iy. This tells us that exp(z) = u(x,y) + iv(x,y), where
u{x,y) - excosy and v(x,y) = -ex
smy.
Suppose that the Cauchy-Riemann equations ux = vyand u
y= -vx are satisfied at some
point z = x + iy. It is easy to see that, for the functions « and v here, these equations become
cos y = and sin y = 0. But there is no value of y satisfying this pair of equations. We mayconclude that, since the Cauchy-Riemann equations fail to be satisfied anywhere, the
function exp(z) is not analytic anywhere.
4. The function exp(z2
) is entire since it is a composition of the entire functions z2and expz;
and the chain rule for derivatives tells us that
-|exp(z2
)=exp(z
2)-|z
2 = 2zexp(z2).
Alternatively, one can show that exp(z2) is entire by writing
exp(z2) = exp[(* + iy)
2
j= exp(*
2 - y2)exp(i2*y)
= exp(x2-y
2)cos(2ry) + iexp(jc
2 - y2)sin(2;ty)
>w
< » „/
U V
and using the Cauchy-Riemann equations. To be specific,
ux =2xexp[x2 -y2
)cos(2;ty)-2yexp(;t2-y
2)sin(2xy) = v
y
and
uy= - 2y exp(x
2 - y2) cos(2xy ) - 2x exp(x
2 - y2)sin(2.xy) = -vx .
36
Furthermore,
-y-exp(z2
)= ux + ivx = 2(x + iy)[exp(*
2 -
y
2)cos(2xy) + iexp(x
2 -
y
2)sin(2xy)]
= 2zexp(z2
).
5. We first write
|exp(2z + i)\ = \exp[2x + i(2y + 1)]|= e
2x
and
|exp(iz2
)|= |exp[-2xy + i(x
2 - y2
)]|= e
2xy.
Then, since
|exp(2z + 1) + exp(iz2
)|< |exp(2z + 1)| + |exp(iz
2
)|
,
it follows that
|exp(2z + 1) + exp(/z2
)|< e
2x + e~2xy
.
6. First write
|exp(z2
)|= |exp[(x + ry)
2
]|= |exp(j:
2 -/) + i2xy| = expU2 -y2
)
and
exp(lzl2) = exp(jc
2 + y2).
Since x2 - y
2 < x2 + y
2, it is clear that exp(*
2 - y2) < exp(*
2 +y2). Hence it follows from
the above that
|exp(z2)|<exp(lzl
2).
7. To prove that |exp(-2z)| < 1 <=> Re z > 0, write
|exp(-2z)| = |exp(-2;c - /2y)| = exp(-2*).
It is then clear that the statement to be proved is the same as exp(-2*) < 1 <=> x > 0, which is
obvious from the graph of the exponential function in calculus.
37
(a) Write el = -2 as eV = 2*'*. This tells us that
e*=2 and y = n + 2nn (n = 0,±l,±2,...).
That is,
* = ln2 and y = (2n + l);r (n = 0,±1,±2,...).
Hence
z = ln2 + (2n + l)«f ' (n = 0,±l,±2,...).
(b) Write *l = 1 + V3 1 as <?V = 2e
,(*/3), from which we see that
e*=2 and y =j + 2nn (n = 0,±1,±2,...).
That is,
* = ln2 and y = ^2» + |j« (n = 0,±1,±2,...).
Consequently,
z = \xi2 + {ln + ^jti (« = 0,±1,±2,...).
(cj Write exp(2z - 1) = 1 as e2x~l
ei2y = le
i0and note how it follows that
<?
2x_l =l and 2y = + 2nx (n = 0,±l,+2,...).
Evidendy, then,
2
and this means that
x = \ and y = n^ (n = 0,±1,±2,...);
z = -^ + nm (n = 0,±l,±2,...).
This problem is actually to find all roots of the equation
exp(iz) = exp(/z).
38
To do this, set z = jc + iy and rewrite the equation as
eye'
a =eye".
Now, according to the statement in italics at the beginning of Sec.8 in the text,
e~y = e
y and -x = x + Inn,
where n may have any one of the values n = 0,±l,±2,.„. Thus
y = and x = nn (n = 0,±1,±2,...).
The roots of the original equation are, therefore,
z=nn (n = 0,±l,±2,...).
10. (a) Suppose that ez
is real. Since el = e
xcosy + ie
xsin y, this means that e
xsiny = 0.
Moreover, since ex
is never zero, siny = 0. Consequently, y = nn(n = 0,±1,±2,...);
that is, Imz = n^(« = 0,±l,±2,...).
(b) On the other hand, suppose that ez
is pure imaginary. It follows that cosy = 0, or that
y = ^- + nn(n = Q,±\,±2,...). Thatis, Inu =— + /i7r(n = 0,±l,±2,...).21 2
12. We start by writing
1 _ z _ z x-iy
z zz Izl2
x 2 +y2x
2 + y2
x2 +y2
x . -y+ i-
'
Because Re(e z) = e* cosy, it follows that
Re(<?1/Z
) = exp2 . 2\x +y
cos
^2 + y2
= exp.2 , ..2
COS2
, ..2jc +y J \x +y
Since is analytic in every domain that does not contain the origin, Theorem 1 in Sec. 25
ensures that Re(e1/z) is harmonic in such a domain.
13. If f(z) = u(x,y) + i'v(jc,y) is analytic in some domain D, then
efit) = e
u(x,y)CQS^ y) + ie
u(*,y)sinv(xy)
Since efU) is a composition of functions that are analytic in D, it follows from Theorem 1 in
Sec. 25 that its component functions
U(x,y) = eu(x
-y)cosvU.y), V(x,y) = e
u(x 'y)sinv(jc,y)
39
are harmonic in D. Moreover, by Theorem 2 in Sec. 25, V(x,y) is a harmonic conjugate of
U(x,y).
14. The problem here is to establish the identity
(expz)" = exp(nz) (n = 0,±1,±2,...).
(a) To show that it is true when n = 0,1,2,..., we use mathematical induction. It is
obviously true when n = 0. Suppose that it is true when n - m, where m is any
nonnegative integer. Then
(expz)m+l = (expz)"
1
(expz) = exp(mz)expz = exp(mz + z) = exp[(m + l)z],
(b) Suppose now that n is a negative integer (n = -1, -2, . .. ), and write m = -n = 1, 2, . . . . Li
view of part (a),
(expz)" =^expz;
1 1 1
(expz)m
exp(znz) exp(-nz)= exp(nz).
SECTION 30
1. (a) Log(-ei) = \n\-ei\+iArg(-ei) = lne-^i = l-^i.
(b) Log(l- = lnll - i1+i'Arg(l - = InV2 — = |ln2 - -ji.
2. fa; loge = lne + i(0 + 2n7f) = l + 2/i7Fi (n = 0,±l,±2,...).
(b) logi = lnl+i^ + 2nnj = ^2n +^m (n = 0,±l,±2,...).
(c) log(-l + V3i) = ln2 +i^ + 2n^ = ln2 + 2^n +|jm (n = 0,±1,±2,...).
3. (a) Observe that
and
Log(l + 1)
2 = Log(2i) = In 2 + ~/
2Log(l + 1) = 2^1nV2 + ijj = In 2 +
Thus
Log(l + i)2 =2Log(l + i).
(b) On the other hand,
Log(-l + 1)2 = Log(-2/) = In 2 - -/
and
2Log(-l + i) = 2( lnV2 +i^J = ln2 + y/.3n
Hence
Log(-l + /)2*2Log(-l + /).
(a) Consider the branch
Since
logz = lnr + /0 r>0, — <6<—4 4
log(i2) = log(-l) = lnl + i;r=7B and 21og/ = 2flnl +
/-|J
= m,
we find that log(/2) = 2 log/ when this branch of logz is taken.
(b) Now consider the branch
Here
logz = lnr + /0 r > 0,— < 9 <4 4
log(i2) = log(-l) = In 1 + in = m and 2 log/ = 2 In 1 + /
,5^2 ,
= 5m.
Hence, for this particular branch, log(/2) * 2 log/.
(a) The two values of im
are e'W4
and ei5tt/
\ Observe that
( Klog(€'*
/4) = lnl + /|-^+ 2«7r| =
( l\2n +- \jti (« = 0,±1,±2,...)
and
log(*,9W4
) = lnl + i|^ + 2/«rh5n
(2n + l) +-4
7n (n = 0,±l,±2,...).
Combining these two sets of values, we find that
log(/1/2
) = |n +^nm (n = 0,±l,±2,...).
41
On the other hand,
1, . 1-logi=-2 2
lnl + i| ^ + 2nn = \n +—\m (n = 0,±l,±2,...).
1Thus the set of values of log(i ) is the same as the set of values of — logi, and we
2may write
log(/1/2
) = |log/.
(b) Note that
log(i2) = log(-l) = In 1 + {it + 2mt)i = {2n + \)iti (n = 0,±l,±2,...)
but that
21ogi' = 2 lnl + i'l ^+ 2mt = {An + Y)Td (/i = 0,±l,±2,...).
Evidently, then, the set of values of log(/2) is not the same as the set of values of
2 log i. That is,
log(i2)*21ogi.
7. To solve the equation logz = iitll, write exp(logz) = exp(/7r / 2), or z = e'*n = i.
10. Since ln(jc2 +y
2) is the real component of any (analytic) branch of 2 logz, it is harmonic in
every domain that does not contain the origin. This can be verified directly by writing
u{x,y) = ln(*2 + y
2) and showing that ^(jc.y) + u (x,y) = 0.
SECTION 31
1. Suppose that Rezt> and Rez2 > 0. Then
zt= r
texp i'0j and z1 -r1
exp i'02 ,
where
it ~ it , it ^ it— < 0, <— and <©,<— ,
21
2 22
2
42
The fact that -% <t+ 2 < n enables us to write
Log(ztz2 ) = Log[(r
xr2 )exp /(0
t+ 2 )] = lnfe) + i{Q
x+ 2 )
= (In r{+ i'0i ) + (In r2 +
1'02 ) = Logfo exp i©, ) + Log(r2 exp i02 )
= LogZ! + Logz2 .
3. We are asked to show in two different ways that
log =logzt-logZ2 (z, * 0, z2
?t 0).
(a) One way is to refer to the relation arg — = argZ[ - argz2 in Sec. 7 and write
log = ln + iarg = (lnl^l+zargz!) - (lnlz2 l+i'argz2 )= log^ - logz2 .
(bj Another way is to first show that log[ -]= -logz (z * 0). To do this, we write z = re
and then
log^ = log^e"'8
j= ln^+ i(-0 + 2imt) = -[lnr + i(0- 2imt)] = -logz,
where n = 0,±1,±2,. . . . This enables us to use the relation
log(z1z2 ) = logz
1+logz2
and write
log
\ Z2J
log = logz, + log
f\2 = logz,- logz2
43
5. The problem here is to verify that
z1/n =exp -UogzJ (n = -l,-2,...),
given that it is valid when n = 1,2,... . To do this, we put m = -n, where n is a negative
integer. Then, since m is a positive integer, we may use the relations z'1 = 11 z and
1/ ez = e~
zto write
= 1,
p(^iog4
exp(llogz)] = exp(-llogz) = expQlogz).
SECTION 32
1. In each part below, n = 0,±l,±2,....
(a) (1 + 1)' = exp[i log(l + 1)] = exp< i
{i1= exp| —ln2-[ — + 2nn
n
lnV2+i(^+ 2n^
= exp^--^ - 2n^jexp^ln2j.
(b)
Since n takes on all integral values, the term -2nn here can be replaced by +2nit.Thus
(1 + /)'' =exp^-| + 2n^exp^ln2j.
(-1)1'* = exp^log(-l)J = exp|^[ln 1 + i(n + 2nn]^ = exp[(2n + 1)/].
2. (a) P.V. /' = exp(/Logi) = exp^ln 1 + /-|
j
exp - n
(b) P.V. [j(-l - V3/)]"= exp|3^Log[|(-l - V3/)| = exp
= exp(2/r2)exp(/3^) = -exp(2;r
2
)
3ni\ )ne-iIn
44
(c) P.V. (1 - i)4
' = exp[4/Log(l - 1)] = exp
= e*[cos(41nV2) + isin(41n V2)] = «*[cos(21n2) + isin(21n2)].
4iflnV2-/j
3. Since - 1 + V3i = 2e2w</3
, we may write
(-1 + V3/)3'2 =exp |log(-l + V30] = exp|| ln2 + /|
^Y+ lnn
= exp[ln(23/2
) + (3n + l)*ri] = 2-^2 exp[(3/i +
where n = 0,±1,±2,.... Observe that if n is even, then 3n+l is odd; and so
exp[(3n + l);n'] = -l. On the other hand, if n is odd, 3n+l is even; and this means that
exp[(3/i + Y)iti] = 1. So only two distinct values of (-1 + V3/)3/2
arise. Specifically,
(-1 + V3i)3/2
=±2V2.
5. We consider here any nonzero complex number z in the exponential form Zq = rQexp i'0
o ,
where -it < © < it. According to Sec. 8, the principal value of zUn
is ^/r^exp^i—j; and,
according to Sec. 32, that value is
exp( —Logz|= exp ^ (in r + i'0
o ) = exp(ln ^/r~) exp^z' = nfc exp^j^
These two expressions are evidently the same.
7. Observe that when c = a + bi is any fixed complex number, where c * 0,±1,±2,..., the
power iccan be written as
T = exp(clogi) = exp< (a + bi) ]sil + i^— + 2nit
= expit
-b — + 2mt \+ia\ — + 2mt.12 ) Vl
it
Thus
(n = 0,±l,±2,...).
I/Cl= exp -b\ — + 2nit (» = 0,±1,±2,...),
and it is clear that lzc
I is multiple-valued unless b = 0, or c is real. Note that the restriction
c * 0,±1,±2,... ensures that ic
is multiple-valued even when b = 0.
SECTION 33
1. The desired derivatives can be found by writing
—sin = ±f'k -'~k ^
dz dz
e--e- = j_(d iz _d- iz
2i )
e-—-e -
j2i\dz dz
and
= Yi\ie +ie )=—
2— = cosz
d d(e* + e-*} lfd k d _fe—cosz =— =- — e +—e
dz dz\ 2 ) 2\dz dz
= -[ie -ie ")- = = -sinz.2 V
i 2i
2. From the expressions
e*-e'* , ek + e~
iz
s'mz = and cosz =2i 2
we see that
ek+e-'
K,ek -e~k kcosz + ismz = + = e .
2 2
3. Equation (4), Sec. 33 is
2 sin ztcos z2 = sinfo + z2 ) + sin^ - Zj ).
Interchanging Zi and z2 here and using the fact that sin z is an odd function, we have
2 cos zxsin z2 = sin(z
t+ z1 ) - sin(zi - z2 ).
Addition of corresponding sides of these two equations now yields
2(sin zxcos z^ + cosz
vsin Zj ) = 2 sin(z
t+ Z2 ),
or
sin(z! + z2 ) = sin z, cos z2 + cos z{sin 22
.
4. Differentiating each side of equation (5), Sec. 33, with respect to zv we have
cos(zt+z2 )
= cosZ[ cosz2 -sinztsinz2 .
46
7. (a) From the identity sin2z + cos
2z = 1, we have
sin2z cos
2z 1 ,22—— +—— =—— ' or 1 + tan z = sec z.
cos z cos z cos z
ft) Also,
sin z cos z 1
. , +. , =
. , , or 1 + cot z = esc z.sm z sin z sin z
9. From the expression
sin z = sinxcosh y + 1 cosxsinh y,we find that
Isinzl2 = sin
2xcosh
2y + cos
2x sinh
2y
= sin2x(l + sinh
2y) + (1 - sin
2jc)sinh
2
y
= sin2x + sinh
2y.
The expression
cosz = cosxcosh y + i sin*sinh y,
on the other hand, tells us that
Icoszl2 = cos
2jccosh
2y + sin
2jtsinh
2y
= cos2*(1 + sinh
2y) + (1 - cos
2x) sinh
2y
= cos2x + sinh
2y.
10. Since sinh2y is never negative, it follows from Exercise 9 that
(a) Isinzl2 > sin
2x, or Isinzl > lsin*l
and that
(b) Icoszl2 > cos
2x, or Icoszl > I cos xl.
11. In this problem we shall use the identities
I sin zl2 = sin
2 x + sinh2y, I cos zl
2 = cos2 x + sinh
2y.
47
(a) Observe that
sinh2y =lsinzl
2 - sin2x < Isinzl
2
and
Isinzl2 = sin
2 x + (cosh2y - 1) = cosh
2y - (1 - sin
2x)
= cosh2y - cos
2 x < cosh2y.
Thus
sinh2y<lsinzl
2 < cosh2y, or lsinhyl<lsinzl< coshy.
(b) On the other hand,
sinh2y = I cos zl
2 - cos2x < I cos zl
2
and
Icoszl2 = cos
2 x + (cosh2y - 1) = cosh
2y - (1 - cos
2x)
= cosh2y - sin
2x <, cosh
2y.
Hence
sinh2y < Icoszl
2^ cosh
2y, or lsinhyl< Icoszl^ coshy.
13. By writing /(z) = sinz =sin(;c- iy) = sin* cosh y -icos* sinh y, we have
f(z) = u(x,y) + iv(x,y),
where
u(x,
y
) = sinx coshy and v(x,y) = -cosxsinhy
.
If the Cauchy-Riemann equations wx = vy ,
Wj, = -v, are to hold, it is easy to see that
cos;ccoshy = and sin*sinhy = 0.
Since coshy is never zero, it follows from the first of these equations that cosx = 0; that is,
x =— + nn (n = 0± 1,±2,...)- Furthermore, since sinx is nonzero for each of these values
2
of x, the second equation tells us that sinhy = 0, or y = 0. Thus the Cauchy-Riemann
equations hold only at the points
z =- + n7t (n = 0±l,±2,...)-2
Evidendy, then, there is no neighborhood of any point throughout which/ is analytic, and
we may conclude that sinz is not analytic anywhere.
The function /(z) = cosz = cos(jc -/y) = cosx coshy + isinx sinhy can be written as
/(z) = u(x,y) + iv(x,y),
where
u(x, y) = cosx coshy and v(x, y) = sin xsinh y.
48
If the Cauchy-Riemann equations ux = vy ,
uy~ -vx hold, then
sinxcoshy = and cosxsinhy = 0.
The first of these equations tells us that sin;c = 0, or x = n7t(n = 0,±l,±2,...). Since
cosnn *0, it follows that sinhy = 0, or y = 0. Consequently, the Cauchy-Riemann
equations hold only when
z = nn (n = 0±l,±2,...).
So there is no neighborhood throughout which / is analytic, and this means that cosz is
nowhere analytic.
16. (a) Use expression (12), Sec. 33, to write
cos(iz) = cos(-y + ix) = cosy coshx — i siny sinhx
and
cos(i'z) = cos(y + ur) = cosycoshx-isinysinh*.
This shows that cos(i'z) = cos(iz) for all z.
(b) Use expression (1 1), Sec. 33, to write
sin(i'z) = sin(-y + ix) = -sinycoshx - icosysinhx
and
sin(i'z) = sin(y + ix) - siny coshx + i cosysinh jc.
Evidently, then, the equation sin(zz) = sin(i'z) is equivalent to the pair of equations
sinycoshx = 0, cosysinhx = 0.
Since cosh* is never zero, the first of these equations tells us that siny = 0.
Consequently, y = nn (n = 0,±1,±2,...). Since coshtt = (-1)" ^ 0, the second
equation tells us that sinhx = 0, or that x = 0. So we may conclude that
sin(/z) = sin(/z) if and only if z = + inn = nni (n = 0,±1,±2,. . .).
17. Rewriting the equation sinz = cosh4 as sin;ccoshy + icos;tsinhy = cosh 4, we see that we
need to solve the pair of equations
sinx cosh y = cosh 4, cosxsinhy =
49
for x and y. If y = 0, the first equation becomes sin* = cosh 4, which cannot be satisfied by
any* since sin*^l and cosh4>l. So y*0, and the second equation requires that
cos* = 0. Thus
x = ~- + n7t (n = 0±l,±2,...).
Since
sin^Y + »wj = (-l)",
the first equation then becomes (-1)" coshy = cosh 4, which cannot hold when n is odd. If n
is even, it follows that y = ±4. Finally, then, the roots of sinz = cosh 4 are
Z =(f
+ 2n7C]±4i (n = 0±l,±2,...).
18. The problem here is to find all roots of the equation cosz = 2. We start by writing that
equation as cosxcoshy-/sin;tsinhy = 2. Thus we need to solve the pair of equations
cos*coshy = 2, sinjrsinhy =
for x and y. We note that y * since cosx = 2 if y = 0, and that is impossible. So the
second in the pair of equations to be solved tells us that sin;c = 0, or that x = nn
(n = 0±l,±2,...). The first equation then tells us that (-l)"coshy = 2; and, since coshy is
always positive, n must be even. That is, x = 2nn {n = 0±1,±2,...). But this means that
coshy = 2, or y = cosh-12 . Consequently, the roots of the given equation are
z = 2n^ + /cosh_12 (n = 0±1,±2,...).
To express cosh"1
2, which has two values, in a different way, we begin with
y = cosh"1
2, or coshy = 2. This tells us that ey + e~
y - 4; and, rewriting this as
(eyf-4(ey
) + l = 0,
we may apply the quadratic formula to obtain ey = 2 ± V3, or y = ln(2 ± V3). Finally, with
the observation that
ln(2-V3) = ln
we arrive at this alternative form of the roots:
(2-V3)(2 + V3)
2 + V3= ln
2 + V3!
= -ln(2 + V3),
z = 2n;r±tln(2 + V3) (n = 0±l,±2,...).
50
SECTION 34
1. To find the derivatives of sinhz and coshz, we write
d . , d(e z -e'z
) 1 d , . _ZN ez + e'
z
and
d , d (
e
z + ez
\ 1 d , , _, x ez -e z
. ,_coshz=_^__j =__(e!+ ,.)=__ =sinlu.
3. Identity (7), Sec. 33, is sin2z + cos
2z = 1. Replacing z by & here and using the identities
sin(z'z) = isinhz and cos(iz) = coshz,
we find that i'
2sinh
2z + cosh
2z = 1, or
cosh2z - sinh
2z = 1.
Identity (6), Sec. 33, is cos^ + Zj) = cosz^oszj -sinzjsinzj. Replacing Zj by iZj and
z2 by iz2here, we have cosf/^ +z2 )]
= cos(/z1)cos(/z2 )-sin(iz1
)sin(iz2 ). The sameidentities that were used just above then lead to
cosh(z1+ z2 ) = coshzj cosh^ + sinl^ sinhz2.
6. We wish to show that
lsinh.*l<lcoshzl< cosh*
in two different ways.
(a) Identity (12), Sec. 34, is lcoshzl2 = sinh
2x + cos
2y. Thus Icoshzl
2-sinh
2 x>0; and
this tells us that sinh2x < Icoshzl
2, or Isinhx\< Icoshzl. On the other hand, since
Icoshzl2 = (cosh
2 x - 1) + cos2y = cosh
2x - (1 - cos
2y) = cosh
2 x - sin2y, we know that
Icoshzl2-cosh
2 x<0. Consequently, lcoshzl2<cosh2
;t, or lcoshzl<coshx.
(b) Exercise U(b), Sec. 33, tells us that lsinhyl<lcoszl< coshy. Replacing z by iz here and
recalling that cosi'z = coshz and iz = ~y + ix, we obtain the desired inequalities.
7. (a) Observe that
sinh(z + m) = = = = = -sinhz.
51
(b) Also,
cosh(z + 7ti) = Lf =—
—
'-z—1— =——— = -- = -coshz.
(c) From parts (a) and (b), we find that
u/- _ sinh(z + ?n) -sinhz sinhztanh(z + /n) =—— - = — = = tanhz.
cosh(z + 7n) -coshz coshz
9. The zeros of the hyperbolic tangent function
t . sinhztanhz =
coshz
are the same as the zeros of sinhz, which are z = nm (n = 0,±1,±2,...). The singularities of
tanhz are the zeros of coshz, or z = ^-j + rucji (n = 0,±1,±2,...).
15. (a) Observe that, since sinhz = / can be written as sinhxcosy+ /coshxsiny = i, we need
to solve the pair of equations
sinhx cosy = 0, coshx siny = 1.
itIf x = 0, the second of these equations becomes siny = l; and so y =— + 2nn
2
(n = 0,±l,±2,...). Hence
(n = 0,±l,±2,...).
7TIf x*0, the first equation requires that cosy = 0, or y =—vmt
(/i = 0,±l,±2,...). The second then becomes (-l)"coshjr = l. But there is no nonzero
value of x satisfying this equation, and we have no additional roots of sinhz = i.
(b) Rewriting cosh z = ^ as coshx cosy 4- isinhx siny = — , we see that x and y must satisfy2 2
the pair of equations
coshxcosy = — , sinhxsiny = 0.2
52
If x = 0, the second equation is satisfied and the first equation becomes1 1 %
cosy = — . Thus y = cos1 — = ±— + 2nn (n = 0,±1,±2,...), and this means that
z = {ln±^7ti (n = 0,±l,±2,...).
If x 0, the second equation tells us that y = nn (n = 0,±1,±2,...). The first then
becomes (-1)" cosh x = But this equation in x has no solution since cosh* > 1 for
all x. Thus no additional roots of coshz = — are obtained.2
16. Let us rewrite coshz = -2 as cosh* cos ;y + isinhjcsin;y = -2. The problem is evidently to
solve the pair of equations
coshxcosy = -2, sinh x siny - 0.
If * = 0, the second equation is satisfied and the first reduces to cos y = -2. Since there
is no y satisfying this equation, no roots of coshz = -2 arise.
If x*Q, we find from the second equation that siny = 0, or y-nn (n = 0,±l,±2,...).
Since cosn^ = (-l)", it follows from the first equation that (-i)"coshx = -2. But this
equation can hold only when n is odd, in which case x = cosh-1
2. Consequently,
z = cosh"1
2 + (2n + l)m (n = 0,±1,±2,...).
Recalling from the solution of Exercise 18, Sec 33, that cosh"1
2 = ±ln(2 + V3), we note that
these roots can also be written as
z = ±ln(2 + V3) + (2n + l)ni (n = 0,±1,±2,...).
SECTION 37
Chapter 4
2. (a) jQ^J*=j(£-l)<*^
T |v2T ir ^.•f ,i V3 /e dt = —- =— cos— + *sin— -1 =— + -;
o L 2/ 1 2/L 3 3 J 4 4'
fcj Since le~**l= e~bx
, we find that
6 IV" T"* 1 1
fe""</f = lim f e
udt = lira = - lim(l - e
-bt) = - when Re z > 0.
L *• Jj=0 * *
3. The problem here is to verify that
\emB
e-Mde =
o
To do this, we write
?s? r when m*n,
27T when m = n.
2* 2w
/ =JeM
e~MdO = je'^'de
and observe that when m#n,
i(m-n)B
i(m — n)
2ic
1 1
i(m-n) i(jn-n)= 0.
When m = n, /becomes
and the verification is complete.
2*
/= \dd = 2n;
4. First of all,
7* 7t
Je(I+,)x
etc =Jexcos* dtc +
1"
J e* sinx dx.
But also,
,(!+<>
l + I
e*e'"-l -e*-l l-i l + e* A + e"+ i—-—
.
l + i l + i l-i
Equating the real parts and then the imaginary parts of these two expressions, we find that
je'cosx dx = - ^+ e
and jVsin;c_* =o
2
Consider the function w(t) = e" and observe that
lie 2tc
jw(t)dt=je it
dt =
l + e*
=i-i=o.i i
Since \w(c)(2n - 0)| =\eK\2n = 2% for every real number c, it is clear that there is no number
c in the interval < t < 2n such that
lie
jw(t)dt=w(c)(2n-0).
(a) Suppose that w(t) is even. It is straightforward to show that u(t) and v(f) must be even.Thus
a a a a a
jw(t)dt=ju(t)dt+ijv(t)dt=2ju(t)dt + 2ijv(t)dt
= 2
a a a
j u(t)dt + z
Jv(t)dt = 2
Jw(t)dt.
(b) Suppose, on the other hand, that w(t) is odd. It follows that u(i) and v(f) are odd, and so
a a a
Jw(t)dt = ju(t)dt +
1J v(t)dt = + / = 0.
Consider the functions
1K _____ "
P„(x) = ~j(x + iVl - x2cos ej d6 (n = 0,1,2,...),
where -1 < x < 1. Since
x + i^l-x2 cosd = V*2 + (1 -
x
2)cos
2 d<^x2 +(l-
x
2) = 1,
it follows that
1 T.| X
P„(*)|<-f Lc + iVl-x2cos0 ^<-fj0 = l.
55
SECTION 38
1. (a) Start by writing
—a -a -a
I =Jw(-t)dt =
Jui-t)dt + i j v(-t)dt.
-b -b -b
The substitution r = -t in each of these two integrals on the right then yields
a a b b b
I = -)u(T)dT-ijv(T)dz = ju(t)dr+ij v(t)dr = jw(r)dt.b b a a a
That is,
-a b
jw(-t)dt = jw(r)dt.
-b a
(b) Start with
b b b
1 = jw(t)dt = ju(t)dt+ijv(t)dt
a a a
and then make the substitution t = <p(x) in each of the integrals on the right. The result
is
P P P
I = ju[<l>(T)]<t>'(T)di:+ ijv[</>(T)]<t>'(r)dr =J
{x)dx.
a a a
That is,
b P
jw(t)dt=jw[(/)(x)]<}>f(x)dx.
3. The slope of the line through the points (a, a) and (fi,b) in the ft plane is
b-am = — .
p-a
So the equation of that line is
b-a ,t-a = - (t-a).p-a
56
Solving this equation for t, one can rewrite it as
b-a aB-bat = — T+-^ .
B-a B-a
Since t = 0(f), then,
... b-a aB-baB-a B-a
4. If Z(t) = z[^(T)] , where zit) = x(t) + iy(t) and t = 0(t), then
Z(T) = 4^(i:)] + /y[0(T)].
Hence
Z'(t) = + i-f =x'MtWW +iy'miWWax at
=vim]
+
wcd = z'[<ktW(t).
5. If w(t) = f[z(t)] and f(z) = u(x,y) + iv(x,y), z(t) = x(t) + iy(t), we have
w(f) = u[x(t),y(t)] + Hx(t),y(t)].
The chain rule tells us that
du , . , dv , ,— = uxx+uyy and — = Vj(x' + v
yy',
and so
w'(0 = (uxx' + uyy') + i(yxx' + v
yy').
In view of the Cauchy-Riemann equations ux= v
yand u
y= -v,, then,
w'(t) = (uxx' - vxy') + i(yxx' + uj) = (ux + ivx )(*' + iy').
That is,
w'(0 = {ux [x(t),y(t)] + ivMt)Mt)])[x'(t) + //(f)] = f'[z(t)]z'(t)
when f = f -
SECTION 40
1. (a) Let C be the semicircle z = 2e'° (0 < < n), shown below.
-2 2 x
Then
= 2/ +e = 2j'(j + # + 1) = -4 + 2m.
(b) Now let C be the semicircle z = 2ei0 (n<d< 2ri) just below
This is the same as part (a), except for the limits of integration. Thus
f£±2<fc = 2/
.'9
+
2k
= 2i(-i + 2n-i-7t) = 4 + 2m.
(c) Finally, let C denote the entire circle z = 2ei0(0 < < 2k). In this case,
f ^-^-dz = Am,JC 7
the value here being the sum of the values of the integrals in parts (a) and (b).
2. (a) The arc is C:z = l + ei0 (n< 6<2n). Then
2jt Ijc
Jc(z - \)dz = J(l + «
w - l)few<*0 = ije
nedO = i
.''29
2/
=I(^_^) = I(1_1)=
.
58
(b) Here C: z = x (0 < x < 2). Then
= 0.
3. In this problem, the path C is the sum of the paths Q , C2 , C3 , and C4 that are shown below.
The function to be integrated around the closed path C is /(z) = neKt
. We observe that
c = ci+ C
2 + C3 +Q and find the values of the integrals along the individual legs of thesquare C.
(i) Since C, is z = x (0 < x < 1),
i
Jc^^ =^^ = ^-1.
1
o
(ii) Since C2 is z = 1 + iy (0 < y < 1),
i i
(Hi) Since C3 is z = (1 - x) + 1 (0 < x < 1),
i i
o
(ivj Since C4 is z = i(l - y) (0 < y < 1),
i i
J *»*<fe =4^^'HMy = ^JV^y = -2.
o
Finally, then, since
we find that
\cne*~
ldz = 4{e* -1).
Jc(z-l)dz = J(jt-l)<£c = -jc
The path C is the sum of the paths
Q:z = x + ix3 (-1<*<0) and C2 :z = x + ix
3(0<x£l).
Using
/(z) = lonC, and f(z) = 4y = 4x3on C2 ,
we have
1
jcf(z)dz = [ f(z)dz +JC2 /(z)<fe
=Jl(l + i3x2)<£c +
J4*3(l + i3*
2)<it
VJ-1 o
1 1
= Jit + 3/jA;2dbc +4j^ + 12ijA:
5dtc-1-10
= Wlli + i[*3
t + [*t + 2i: [*
6
]J= 1 + i+ 1 + 2/= 2 + 3/.
The contour C has some parametric representation z = z(t) (a<t< b), where z(a) = zxand
z(fc) = Z2. Then
Jc<fe = \z\t)dt =
[Z(0]* = z(b) - z(a) = z2 -zva
To integrate the branch
around the circle C: z = e,e(0 < < 2;r), write
Let C be the positively oriented circle 1*1=1, with parametric representation
z = e,e
(0 < B < lit), and let m and n be integers. Then
lczmz"dz = j
{ei6
)
m(e-
ie
)
n
i£edd = i je'<"
+1)VM</0.
But we know from Exercise 3, Sec. 37, that
lK[0 when m*n,
[lit when m-n.jeM
e-Md0--
60
Consequently,
frt\fc-j°when m+1 * n -
lc \2m when m + l = n.
8. Note that C is the right-hand half of the circle x2 + y
2 = 4. So, on C, * = ^4-y2. This
suggests the parametric representation C:z = ^4-y2+iy (-2 < y < 2), to be used here.
With that representation, we have
dy
2 2 ( %
= j(-y + y)dy + ij\-tI= +^y2
-2 iW4- ^
2 ..2 4 = 4i
= 4/ [sin"1
(1) - sin"1
(-1)] = 4i = Am.
J -2
10. Let C be the circle z = z + Re ie (-Jt<6< n)
(a) f=
f-^Rie iedd = i]dd = 2m.
-K
(b) When n = ±1,±2,...,
=— (em* - e-'"
1
*) = /— sinn/r =2R° .
n
11. In this case, where a is any real number other than zero, the same steps as in Exercise 10(b),
with a instead of n, yield the result
f (z-Zo)a~ l
dz =i—-sin(a^)Jca a
61
12. (a) The function f(z) is continuous on a smooth arc C, which has a parametric
representation z = z(t) (a<t< b). Exercise 1(b), Sec. 38, enables us to write
b p
\f[z(t)]z'(t)dt =f/[Z(T)k'[0(T)]0'(T)</T,
a a
where
Z(T) = Z[0(T)] («<T<j3).
But expression (14), Sec 38, tells us that
zWt)W'(t) = Z'(t);
and so
* P
jf[z(t)]z\t)dt =jf[Z(r)]Z'(t)dr.
(b) Suppose that C is any contour and that f(z) is piecewise continuous on C. Since C can
be broken up into a finite chain of smooth arcs on which f(z) is continuous, the
identity obtained in part (a) remains valid.
SECTION 41
1. Let C be the arc of the circle \z\= 2 shown below.
o 2 x
Without evaluating the integral, let us find an upper bound forjjc~T~J
note that if z is a point on C,
|z2-l|>|lz
2l-l| = |lzl
2-l| = l4-ll = 3.
To do this, we
Thus
62
Also, the length of C is ^(47r) = n. So, taking M = ^ and L = n , we find that
2. The path C is as shown in the figure below. The midpoint of C is clearly the closest point on
V2C to the origin. The distance of that midpoint from the origin is clearly -y-, the length ofC
being V2.
V2Hence if z is any point on C, Id >— . This means that, for such a point
Consequendy, by taking M - 4 and L = V2 , we have
Izl4
||c^|S ML = 4V2.
3. The contour C is the closed triangular path shown below.
To find an upper bound for\jc
(ez - z)dz , we let z be a point on C and observe that
\ez -
z
I
<
\el
\ + \z\ = ex + jx2 +y2
.
63
But ex < 1 since x < 0, and the distance ^x2 + y
2of the point z from the origin is always
less than or equal to 4. Thus le* - zl < 5 when z is on C. The length of C is evidently 12.
Hence, by writingM = 5 and L = 12, we have
(ez-z)dz < ML = 60.
Note that if lzl= R (R > 2), then
l2z2-ll<2lzl
2+l = 2/?
2 +l
and
lz4+ 5z
2+ 41 = lz
2+ II lz
2 + 41 >|Izl
2-1
1
1 Izl2-4 | = (tf
2 - 1)(/?2 - 4).
Thus
2z2 -l
z4+5z
2 +4I2z
2-ll 2*2 + l
z4 + 5z
2+4l (R
2-l)(R2 -4)
when lzl= R (R > 2). Since the length of CR is ^R, then,
I,
2z2 -l
W+5z2 + 4
^ teR(2/?2 + 1)
~(/?2-l)(/?
2 -4)/?l R2
R2J} R2
)
and it is clear that the value of the integral tends to zero as R tends to infinity.
Here CR is the positively oriented circle Izl=R(R> 1). If z is a point on CR , then
Logz lln/? + /0l< ln/?+l0l
<^r + ln/g
?2 " R2
since -n<Q<7t. The length of Q is, of course, 2nR. Consequently, by taking
M =——— and L = 2kR,
we see that
Since
it follows that
Logzdz
* z< ML = 2n
n + k\R
R
it + \xiR _ 1//?iim = lira —-— = u,
J c* z
Let Cpbe the positively oriented circle \z\=p (0 <p < 1), shown in the figure below, and
suppose that /(z) is analytic in the disk Id < 1.
y
I I' /a 1 \
1 V Jppx
We let z represent any particular branch
-1/2 = exp^-^log zj= exp|^-^ (In r+ id) = exp^-i-j
|(r > 0, a < < a + 2ft)
of the power function here; and we note that, since /(z) is continuous on the closed
bounded disk Izl < 1, there is a nonnegative constantM such that l/(z)l^M for each point z
in that disk. We are asked to find an upper bound for
observe that if z is a point on Cp ,
M
I zmf{z)dz . To do this, we
\z-U2m = \z-
m\\az)\<
Since the length of the path Cp is 2np, we may conclude that
jcz-mf(z)dz <^j=2np = 27tMjp.
Note that, inasmuch as M is independent of p, it follows that
limf z-U2f(z)dz = 0.
65
SECTION 43
1. The function z" (n = 0,1,2,...) has the antiderivative z"+1/(n + 1) everywhere in the finite
plane. Consequently, for any contour C from a point zx to a point z2 ,
.Zl _»+l 1
Jc^ n + 1 . n + 1 /i + l n + 1^2 1
'
"r2
2. fa; I e*z<fe
=
—
i/2
e —e j + 1 l + i
it it it
(b)J
cosl — l<fe = 2sin
>r+2i
= 2 Si„(f + ,)=:2/
= —lie e —e e I
=~<H=7 + « = « + -.e
U-2)41
3
1 1
4i
4 40.
3. Note the function (z -Zq)"'1
(n = ±1,±2,...) always has an antiderivative in any domain that
does not contain the point z = Zq. So, by the theorem in Sec. 42,
L(z-zor l
dz =
for any closed contour C that does not pass through Zq.
5. Let C denote any contour from z = -1 to z = 1 that, except for its end points, lies above the
real axis. This exercise asks us to evaluate the integral
t
I = jz'dz,
-i
where z' denotes the principal branch
z' = exp(/Logz) (lzl>0,-^< Argz< it).
66
An antiderivative of this branch cannot be used since the branch is not even defined at
z = -1. But the integrand can be replaced by the branch
z' = exp(z'logz) lzl>0,-— <argz<—
since it agrees with the integrand along C. Using an antiderivative of this new branch, wecan now write
_('+!
1 + 1
- *[(l)
i+l
(j- ^
j~£('>D'ogl
£(M)\og(-D
j
_ _£_("^(i+l)(Inl+iO) _ (i+lXlnl+i*)"! = _J_ h _ g-*^*) - ^ + e1 . 1 L
i + Ji + V ' l + i 1-/
d-o.
SECTION 46
2. The contours Ctand C2 are as shown in the figure below.
In each of the cases below, the singularities of the integrand lie outside Cxor inside C2 ; and
so the integrand is analytic on the contours and between them. Consequendy,
imdz=if(z)dz.
67
(a) When f(z) = - the singularities are the points z =
z + 2(b) When f(z) = . .
—, the singularities are at z = 2nn (n = 0,±1,±2,...).
sm(z 1 2)
(c) When /(z) = -, the singularities are at z = 2nm (n = 0,±1,±2,. . .).
(a) In order to derive the integration formula in question, we integrate the function earound the closed rectangular path shown below.
Since the lower horizontal leg is represented by z = x (-a < x < a), the integral of
e~l* along that leg is
je-xldx = 2je-
xldx.
-a
Since the opposite direction of the upper horizontal leg has parametric representation
z = x + bi (-a £x<a), the integral of e'1* along the upper leg is
-]e'{x+bi)1
dx = -ebl\e
xle
iUxdx =V j<f cos2bxdx + ie"
2
\exlsm2bxdx,
or simplyu
-2/2
jy*2
cos2fct<£t.
Since the right-hand vertical leg is represented by z = a+ iy (0 < y < b), the integral of
e~'* along it is
O B
je-(a+iyf
idy = !«-* jeyl
e-i2ay
dy.
68
Finally, since the opposite direction of the left-hand vertical leg has the representation
z = -a + iy (0 < y < b), the integral of e~l* along that vertical leg is
b b
-feH-a+iy)1
idy = -ie-al
jey*e
i2aydy.
According to the Cauchy-Goursat theorem, then,
a a b b
2\e~xldx - 2e
b\je-
xlcos2bxdx + ie** \ey\-'lla
>dy - ie~ajey
*enay
dy = 0;
V_:
f
and this reduces to X4
]e~xlcoslbxdx = e~
b\e~
xldx + e
-(a^bl)]e
yZ
sinlaydy.
(b) We now let a -» °° in the final equation in part (a), keeping in mind the knownintegration formula
\e'xl
dx
and the fact that
e-(al+bl)
\eyl
sinlaydy < e-(a2+bl)
jeyl
dy -> as a -> oo.
The result is
exlcos2bxdx = ^-e-b2
2(b>0).
6. We let C denote the entire boundary of the semicircular region appearing below. It is madeup of the leg C, from the origin to the point z = 1, the semicircular arc C2 that is shown, and
the leg C3from z = -1 to the origin. Thus C = C
t+ C
2+ C
3.
y
69
We also let f(z) be a continuous function that is defined on this closed semicircular regionby writing /(0) = and using the branch
/(z) = V7eie/2
r>o,-f<*<f
of the multiple-valued function zm
. The problem here is to evaluate the integral of f(z)around C by evaluating the integrals along the individual paths C„ C2 ,and C3 and thenadding the results. Li each case, we write a parametric representation for the path (or arelated one) and then use it to evaluate the integral along the particular path.
(i) Cv
: z = rei0 (0<r<l). Then
JCi/(z)^=jVF.lJr =[V^=|.
(ii) C2
: z = 1 • e'B(0 < 6 < it). Then
I f(z)dz = ]ewn
ieiedd = ije^nd0 = if^e
na'2]*= - 1) = -1 (1 + /).
2o o L3i Jo 3 3
(Hi) -Cy z = reix
(0 < r < 1). Then
\cmdz = -\_c
f{z)dz = -)^e i
*'\-\)dr = i\^dr = i\-rvll
\
=-/.5 *" J
o o L3 Jo 3
The desired result is
jcf(z)dz = j f(z)dz + j f(z)dz+i f(z)dz =l-h+i) +h = 0.
1 Cj "-J 3 3 3
The Cauchy-Goursat theorem does not apply since f(z) is not analytic at the origin, or even
defined on the negative imaginary axis.
SECTION 48
1. In this problem, we let C denote the square contour shown in the figure below.
y
2i
ci
-2,r 2 Jt
-2/
(a) f
6 =2m \e~z\ .
, = 2m(-i) = 2n.
(b)f^- dz = f^ 7 + 8>
<fe =iJ^LJc z(z
2 + 8) Jc z-0 Lz2 + 8. U; 4
Jc 2z + 1Jc z -(_i/2) L2J z=_1/2 I 4j 2
. ,. r coshz , r coshz , 2tb* f d3
,
<d> Jc—* Jcb^sr*"IT 60,1,1
Jz=0
=f(0)
= 0.
, . r tan(z/2) , r tan(z/2) . 2«f
Z=*o
= 2«i"| 4-sec2— 1 = »rsec
2f— I when -2<xn <2.U 2
Let C denote the positively oriented circle \z - il= 2, shown below.
(aj The Cauchy integral formula enables us to write
dz c dz r l/(z + 2i)
JCr2 +4 J<z2 +4 Jc(z _2/)(z + 20 Jc z-2/
dz = 2mz + 2ij z=2i
= 2m\
(b) Applying the extended form of the Cauchy integral formula, we have
f& _ f ffc r 1/ U
Jc(z
2 + 4)2 ~ Jc
(z _ 2/)2
(z + 2i)2 ~ Jc (* -
dz _ f l/(z + 2Q2
= 2£(z-2i)
1+1Z
1!
1
<fe (z + 2i)
71
3. Let C be the positively oriented circle Iz I= 3 , and consider the function
g(w)=\ dz (lwl*3).
We wish to find g(w) when w = 2 and when \w\ > 3 (see the figure below).
y
Cm
)( i
V w = 2/3 x
We observe that
#(2) = Jc z_2~ <fe = 2 ^'[2z2 " z ~ 2
] z=2= 2^1(4) = 8m.
On the other hand, when Iwl > 3, the Cauchy-Goursat theorem tells us that g(w) = 0.
5. Suppose that a function / is analytic inside and on a simple closed contour C and that z is
not on C. If z is inside C, then
f£&*»2«rk) and f^>* f/ft)* =^
Thus
f f\z)dz = f /(g)<fe
J^ Z-Zo J^(Z-Z )
2'
The Cauchy-Goursat theorem tells us that this last equation is also valid when zq is exteriorto C, each side of the equation being 0.
7. Let C be the unit circle z = eie (-n£d< it), and let a denote any real constant. The
Cauchy integral formula reveals that
f —dz=[ -^-dz = 2m\e at
]= 2m.
Jc z Jc z — J *=0
On the other hand, the stated parametric representation for C gives us
\c—dz = ]^~Qiewde = / jexp[a(cos0 + isin 6)]d6Z -K e
-It
It K
= iJeacosVflSin ed0 = ije
acose[cos(asm 6) + isin(asin 6)]dd
-ft -It
K It
= - jeacos0sin(asin 8)dd + ije
acos9cos(asin 6)dd.
-It -1t
r eaz
Equating these two different expressions for the integral I— dz, we have
c zIt It
- jeacosesin(asin d)dd + i je
acosBcos(asin d)dd = 2m.
-It -It
Then, by equating the imaginary parts on each side of this last equation, we see that
it
jett<:ose
cos(asmd)dd = 27f,
-«
and, since the integrand here is even,
jeacose
cos(asin 6)dd = Jt.
(a) The binomial formula enables us to write
We note that the highest power of z appearing under the derivative is z2", and
differentiating it n times brings it down to z". So Pn (z) is a polynomial of degree n.
(b) We let C denote any positively oriented simple closed contour surrounding a fized point
z. The Cauchy integral formula for derivatives tells us that
rtf-tf^fSfzlLj, („ = 0,1,2,...).
dz"K
' Im^is-z)^1
Hence the polynomials Pn (z) in part (a) can be written
73
(c) Note that
C?2-l)" _ (j-1)"(j + 1)" +
(s-iyn+l
(s-iy S-l
Referring to the final result in part (b), then, we have
Also, since
(s2 - 1)"
= (s - + 1)" _ (s - 1)"
(j + ir+1
(* + l)"+1
5 + 1'
we have
2" 2to* jc s + i
9. We are asked to show that
1 rf(s)ds
m J (t —ni Jc (s-z)
3*
(a) In view of the expression for /' (z) in the lemma,
/'(z + Az)-/'(z) _ 1 r
Az 2ml1 1
(.y-z-Az)2
(s-z)2
f(s)ds
Az
1f 2(s-z)-Az
2ni[(s-z- Az) (s-z)
Then
/'(z + Az)-/'(z) 1f/(j)<fe _ 1
Az
f JV)as i r
7ri|(5-z)3 2^4
2(j-z)-Az
(s-z-Az)2(s-z)
2(s-z)
3
1f 3(j-z)Az-2(Az)
2
2iTi J f.c —2^/J(,-z-Az)2(5-z)
3"/(^'
74
(b) We must show that
r3(.s-z)Az-2(Az)
2
c(s-z-Azf(s-z?jf(s)ds < (3DIAzl+2IAzl
2 )M(rf-IAzl)V
Now D,d,M, and L are as in the statement of the exercise in the text. The triangle
inequality tells us that
I3(j - z)Az - 2(Az)2l< 31* - zl I Azl + 21 Azl
2 < 3D! Azl + 21 Azl2
.
Also, we know from the verification of the expression for /' (z) in the lemma that
Is - z - Azl > d -I Azl> 0; and this means that
I (s - z - Az)2(s - z)
3l > (d - 1 Azl )
2d3 > 0.
This gives the desired inequality.
(c) Ifwe let Az tend to in the inequality obtained in part (b) we find that
This, together with the result in part (a), yields the desided expression for /"(z).
Chapter 575
SECTION 52
1. We are asked to show in two ways that the sequence
(-1Yz„=-2 +A-f (n = l,2,...)
n
converges to -2. One way is to note that the two sequences
xn =-2 and y„=t-9-(n = l,2,...)
of real numbers converge to -2 and 0, respectively, and then to apply the theorem in Sec.
51. Another way is to observe that \z„- (-2)1 = Thus for each e>0,
z„ ~ (- 2)| < e whenever n>n ,
1where n is any positive integer such that n >
2. Observe that if z„ = - 2 +i^- (n = 1, 2,. .. ), then
n
/•„=lzJ=i/4 +— -»2.
But, since
2fl=Argz
2„-»^ and2„_,
= Argz^, (n = 1,2,...),
the sequence 0„ (n = 1,2,...) does not converge.
3. Suppose that limzB = z. That is, for each e >0, there is a positive integer such that
\zn - z\< £ whenever n > n . In view of the inequality (see Sec. 4)
lzn -zl>llzj-idl,
it follows that llzj-lzlke whenever n>nQ . That is, limlzj=lzl.
76
4. The summation formula found in the example in Sec. 52 can be written
when lzl<l.
If we put z = where < r < 1, the left-hand side becomes
jr(re'*)" = jr rV"e = j^V cosnfl + rnsinn0;
u=l n=l n=l n=l
and the right-hand side takes the form
rew
1 - re~w_ re
w -r2 _ rcos - r2 + irsin
1 - reie
'
1 - re~w ~
1 - r(eie + e~
ie
) + r2 ~
1 - 2rcos + r2
Thus
^ . - . . _ rcos0-r2. rsin0
> r cosnd + iy r smnd = —T + i—T .
Zl t& l-2rcos0 + r2 l-2rcos0 + r
2
Equating the real parts on each side here and then the imaginary parts, we arrive at the
summation formulas
E„ Q rcosd-r2
j V « • a rsin0r cosn0 = —
-r and > r smnd = -— —T ,
B=1 l-2rcos0 + r2 ^ l-2rcos0 + r
2
where < r < 1. These formulas clearly hold when r = too.
6. Suppose that ^zB=5. To show that ^z„ =S, we write zn =xn +iy„, S = X + iY and
n=l n=l
appeal to the theorem in Sec. 52. First of all, we note that
5>„=X and 5>B=7.
OO
Then, since ^(—y„) = —Y, it follows that
jU = 2ta -% )= 2>„ + «*(->.)]
=
x - iY = j.
n=l
77
8. Suppose that £z„ = S and ^w„ = T. In order to use the theorem in Sec. 52, we writeb=1
z» =*,+%> S = X + iY and w„=un +ivn , T = U + iV.
Now
Since
2>b = * f,yH=Y and 5>„ = t/, |>B
= V.B=l B=l B=l
2(^+"„) = X + t/ and X^+vJ^+y,
it follows that
That is,
or
J,[(xn +u„) + i(yn +v„)] = X + U + i(Y+ V).
+%) + ("„ + J'vJ] = X + iY+(t/ + /V),
B=l
B=l
SECTION 54
1. Replace z by z2in the known series
coshz =X~ z
ln
to get
cosh(z2) =2
«> _4bZ
Then, multiplying through this last equation by z, we have the desired result:
zcosh(z2) = ]£
**z4n+1
tS(2n)!
(Izl< ~)
(lzl< ~).
(Izl<~).
78
2. (b) Replacing z by z - 1 in the known expansion
: zn
e = 5>7 <W<«).n=0 " ;
we have
n=0
So
~Z „, (lzk°o).n!
^rt =e£(£zi)l (ld<eo)e
r.=0
3. We want to find the Maclaurin series for the function
z z 1
z4 +9 9 l + (z
4 /9)'
To do this, we first replace z by -(z4/ 9) in the known expansion
r-^ly Ozl<l),
as well as its condition of validity, to get
Then, if we multiply through this last equation by |-, we have the desired expansion:
n
6. Replacing z by z2in the representation
Zo (2n + D!
we have
2n+l
sinz =z
(lzk~),
j
4b+2
sin(z2) = £(-1)" * "„ Qz\< oo).
Since the coefficient of z" in the Maclaurin series for a function f(z) is /(n)
(0)/n!, this
shows that
f4n)(0) = and /
<2B+1)(0) = (n = 0,1,2,...).
7. The function —^— has a singularity at z = 1. So the Taylor series about z = i is valid whenl-z
\z - i\ < V2, as indicated in the figure below.
y
1 i i
\V2 J
To find the series, we start by writing
1 1
l-z (l-i)-(z-O l-i l-(z-/)/(l-0"
This suggests that we replace z by (z - 1) / (1 - i)in the known expansion
1 * n=0
and then multiply through by :• The desired Taylor series is then obtained:
(ld<l)
i-z sa-o*(z-iT.
(lz-/I<V2).
9 The identity sinh(z + ni) = -sinhz and the periodicity of sinhz, with period 2m, tell us that
sinhz = -sinh(z + ni) = -sinh(z - ni).
So, ifwe replace z by z - ni in the known representation
80
and then multiply through by -1, we find that
sinhz = -V i£_JEL!—( | Z _ m\< ~).
S (2» + D!
13. Suppose that < Izl < 4. Then < Iz / 41 < 1, and we can use the known expansion
7^- =2y (IzKl).
To be specific, when < Izl < 4,
1 111 ^fz-lym =y£_ = J_ +y^_ = _L + y.4z-z
24z !_£ 4z~W ^4fl+l
4z ~4"+14Z B
4"+2'
4
SECTION 56
1. We may use the expansion
2n+l
sin-S (-ir™ (UI< ~>
to see that when 0< Izl < °°,
%.j n y (-!>• 1, , f (-ir 1
3. Suppose that 1 <lzl< 00 and recall the Maclaurin series representation
T^- =Z^ (|Z|<1)-1-z ±!'2 „=o
This enables us to write
1
-1 'r-iiRT-i^
1 + Z ^ 1h— Z n=0 v „=o z
Z
Replacing n by n - 1 in this last series and then noting that
81we arrive at the desired expansion:
1 v
4. The singularities of the function /(z) = are at the points z = and z = 1. Hence
there are Laurent series in powers of z for the domains <lzl< 1 and 1 <lzl< <» (see thefigure below).
To find the series when < \z\< 1, recall that =Y z" (lzl< 1) and write
2 A Z Z „=0 n=0 Z Z „=2 „_ Z Z
As for the domain 1 < lzl< <», note that II / zl < 1 and write
/(z)="7'I^17I)
="7|(z")
=~%^~%7-
z + l5. (a) The Maclaurin series for the function is valid when \z\< 1. To find it, we recall
z-1the Maclaurin series representation
TIT = 5>" (ld<1)
for and write1-z
z + l 1~
fzi-<«+ »izr <-< -»X --X^1
- X*'•° A 12 »=0 n=0 «=0
"=1 «=0 11=1
(b) To find the Laurent series for the same function when l<lzl<<*>, we recall the
Maclaurin series for —— that was used in part (a). Since1 - z
< 1 here, we may write
z z
1+-ZJn=i)\ZJ n=oZ n=0 Z
(l<lzl<oo).
The function /(z) =1
has isolated singularities at z = and z = ±z, as indicated inz(l + z
2)
the figure below. Hence there is a Laurent series representation for the domain <lzl< 1
and also one for the domain 1 <lzl< °°, which is exterior to the circle lzl= 1.
To find each of these Laurent series, we recall the Maclaurin series representation
1 2 n=0
(lzl< 1).
For the domain <Id< 1, we have
/(*> = ~'7~~t =1tHT =tc-DV- 1
= -+ £(-1)V" 1 = £(-l)n+1z2n+l
Z 1 + Z Z B=0 B=0 Z Z
On the other hand, when 1 <lzl< °°,
11 1 ^ r__l_V _^ (-If _v (-1)"+1
z2
.2n+l'
n=0 £ n=l£~
In this second expansion, we have used the fact that (-1)"" 1 = (-l)fl-1
(-l)2 = (-1)"
+1.
83
8. (a) Let a denote a real number, where -1 < a < 1. Recalling that
»=0
enables us to write
a a 1
j— = |>" (IzKl)
a _ a i _v fl
z-a"7"l-(a/z)"^7Tr '
or
a ^ a"= 2j— (lal<ld<°o).
(b) Putting z = e'° on each side of the final result in part (a), we have
a n -inBe
But
a (cos0-a)-isin0 acos0-a2-iasin0
eie -a (cos0-a) + i'sin0 (cos0-a)-i'sin0 l-2acos0 + a
2
and
XflV"9 cosn6-i^ansmne.
n=l b=1 n=l
Consequendy,
2,a"cosnd =-— r- and Ya" smn9 = rl-2acosd + a2 ~ l-2acosd + a
2
when -1 < a < 1.
10. (a) Let z be any fixed complex number and C the unit circle w = e'* {-it <<t>^ n) in the w
plane. The function
/(w) = exp2 v w>
has the one singularity w = in the w plane. That singularity is, of course, interior to
C, as shown in the figure below.
w plane
Now the function f(w) has a Laurent series representation in the domain <Im>I< <
According to expression (5), Sec. 55, then,
[IIexp
where the coefficients J„(z) are
(0<lwl<<»),
exp w ——]
2m Jc Wn+l dw (n = 0,±l,±2,...).
Using the parametric representation w = e'* (-n <$<n) for C, let us rewrite
this expression for Jn (z) as follows:
1 r
exp| 2 v~
J" i-z) = T7; J JC»2m ' ev
- K
1*
- iV'd0 =|exp[izsin <l>]e~
m*d<t>
.
2m _
That is,
1K
J„ (z) =^ Jexp[-i - z sin 0)] <fy (n = 0,±l,±2,...).
The last expression for /„ (z)in part (a) can be written as
1*
^i(^)= T~ |[cos(n0-zsin0)-ism(n0-zsin0)]d0
2;r J
\cos(n<t>-zsin<j))d<l> f sin(n</>-z sin </>)d<j>
2tc j„ 2?r j
1 *i
=— 2 1cos(«0 - zsin <j>)d<j>
2% i 2k(h = 0,±1,±2,...).
85
That is,
1*
J„(z) =—Jcos(«0 - zsin
iti(n = 0,±l,±2,...).
11. fa) The function f(z) is analytic in some annular domain centered at the origin; and the
unit circle C.z- e'* (-it < <j> < it) is contained in that domain, as shown below.
For each point z in the annular domain, there is a Laurent series representation
n=0 n=l <
where
— ?T -*?T
(n = 0,1,2,...)
and
— 7F—
Substituting these values of a„ and b„ into the series, we then have
or
/(a.J-J/e-w.J-lJ/,,.,
86
(b) Put z = e'e
in the final result in part (a) to get
or
/(«") = ^- J/(e* )d* +1£ J/(^)cos[n(0 -
If k(0) = Ref(e'e), then, equating the real parts on each side of this last equation yields
w(0) =— J«(0)<ty+—2 J"(0)cos[n(0-0)]<fy.
SECTION 60
1. Differentiating each side of the representation
1-z n=0
we find that
U z; az „=0 n=0 az „=1 n=0
Another differentiation gives
(i — z; "Z n=0 „=o «z B=1 n=0
2. Replace z by 1 / (1 - z) on each side of the Maclaurin series representation (Exercise 1)
as well as in its condition of validity. This yields the Laurent series representation
4 = £<=!>>ZI> «<lz-ll<~).
87
3. Since the function /(z) = 1 / z has a singular point at z = 0, its Taylor series about z = 2 is
valid in the open disk \z-2\<2, as indicated in the figure below.
To find that series, write 1111z 2 + (z-2) 2 l + (z-2)/2
to see that it can be obtained by replacing z by -(z - 2) / 2 in the known expansion
1
Specifically,
or
l_ly|- (Z-2)Tz ifA. 2 J
Z R-0 z
(lzl< 1).
(Iz-2I<2),
(lz-2l<2).
Differentiating this series term by term, we have
J =l^n(z -2r l =E^tH" + l)(z - 2)" (lz - 2I< 2).71 J-~l y
n=0
Thus
-| (_1)>+1)^y (lz-2l<2).
4. Consider the function defined by the equations
V-lf(z) =
1
when z * 0,
when z = 0.
When z * 0, /(z) has the power series representation
/(*) = -z
' 2 3 \
l +£ + l. + £.+.^1! 2! 3!
-1 , z z= 1+—+—+••
2! 3!
Since this representation clearly holds when z = too, it is actually valid for all z. Hence/is entire.
Let C be a contour lying in the open disk \w - II< 1 in the w plane that extends from the
point w = 1 to a point w = z, as shown in the figure below.
w plane
According to Theorem 1 in Sec. 59, we can integrate the Taylor series representation
1
w „=0
term by term along the contour C. Thus
dw
(lw-u<i)
" n=0 n=0
But
and
f— = [— = [Logw]
1
z = Logz-Logl = Logz
4
J(w-l)"=J(w-l)"rfw =
Hence
^n + l ^ n
and, since (-1)"-1 = (-ly-^-l)
2 = (-1)"+1
, this result becomes
Logz =XLii— («-!)"
n + l
(-D
n+l
Jl
(z-1)
n + l
n+l
(Iz-lkl);
(Iz-lkl).n=l
SECTION 61
1. The singularities of the function /(*) =-^L- are at z = , ± L The problem here is to
find the Laurent series for/that is valid in the punctured disk <ld< 1, shown below.
89
yi
\°
4
-i
/ x
We begin by recalling the Maclaurin series representations
and
which enable us to write
and
1! 2! 3!
1
l-z= l + z + z
2 + z3+---
ez
= l + z + ^-z2+ -z
3+-2 6
1
Qz\< oo)
Oz\<l),
(\z\< oo)
(\z\<l).
Multiplying these last two series term by term, we have the Maclaurin series representation
2 ,
- = l + z + ±z2+*z 3+-
zl + \ 2 6
-z2
-z3-
1.2 5
z4+-
which is valid when \z\< 1. The desired Laurent series is then obtained by multiplying each
side of the above representation by -:
1 , 1 5 2= -+l--z
—
z +••z(z' + l) z 2 6
(0<kkl).
90
4. We know the Laurent series representation
1 1117z2sinhz z
36 z 360
Z (0<lzkw)
from Example 2, Sec. 61. Expression (3), Sec. 55, for the coefficients b„ in a Laurent series
tells us that the coefficient of - in this series can be written
z
dz
2m lc z sinhz
where C is the circle lzl= 1, taken counterclockwise. Since bx= - \, then,
6
'c z sinhz
6. The problem here is to use mathematical induction to verify the differentiation formula
(n[/(^(z)]
(n) =E Afk\z)g
"-k)(z) (n = l,2,...).
The formula is clearly true when n = 1 since in that case it becomes
[/(z)s(z)]' = /(z)s'(z) + f(z)g(z).
We now assume that the formula is true when n = m and show how, as a consequence, it is
true when n = m + 1. We start by writing
UWz)](m+1)={[f(z)g(z)]'}
(m)=[f(z)g
,(z) + nz)g(z)]
(m)
=mz)g'(z)fm)+[f(z)g(z)]
l(m)
=1 r /(t)(^(m-t+i)
(^)+s*=0 V* /
/(t)
(z)^(m-t+1)
(z)
= /(z)*(m+1)
(z) +£*=1
'm^ ( m N
/W(z)^('»
+1-^(z) + /
('n+1)(z)^(z).
But
frrC
k ,
m ml|
ml (ro + 1)!
k-l) kl(m-k)l (*-l)!(m-* + l)! kl(m + l-k)l
'm + V
<k
>
and so
minz)g(z)r
+i)=nz)g°
n+i\z)+
y
z
or
[/(^(z)](m+i)
=iTm
,
+rf\z)t«-k
\z).
The desired verification is now complete.
We are given that /(z) is an entire function represented by a series of the form
f(z) = z + a2z2+ajz
3+-'- (lzl<°°).
(a) Write g(z) = f[f(z)] and observe that
It is straightforward to show that
/(z) = /'[/(z)]/'(z),
g"(z) = /"[/(z)][/'(z)]2+ f'[f(z)]f"(z),
and
g"\z) = /'"[/(z)][/'(z)]3
+ 2f(z)f"(z)f"[f(z)] +ft/(z)]/'(z)/"(z)+ /'[/(*)]/"'(*) •
Thus
S(0) = 0, ^'(0) = 1, g"(0)=4a2 , and g'"(0) = 12(a22 + a
3 ),
and so
/[/(z)] = z + 20.Z2+ 2(4 + a,)z
3+- • (lzl< °°).
92
(b) Proceeding formally, we have
/[/(*)] = f(z) + fljC/Cz)]2+ «,[/(z)]
3+- •
•
= (z + fljZ2 + OjZ
3+•••) + (<*2Z
2 + 2^z 3+- • • ) + (fl
3z3+• • •
)
= z +2^ + 2(a* + a3>z3+- • •
.
(c) Since
sinz = z -—+• • •= z + Oz2 + f- -V+- •
•
3! V 6)
the result in part (a), with Oj = and a3= -— , tells us that
6
sin(sinz) = z-— z3 +---
(IzKoo),
(lzl< ~).
8. We need to find the first four nonzero coefficients in the Maclaurin series representation
— =E^z" (\z\<Zcoshz t^n\ \ 2
This representation is valid in the stated disk since the zeros of coshz are the numbers
z = \— + n7t\i (/i = 0,±l,±2,...)» the ones nearest to the origin being z = ±^-i- The series
U J 2
contains only even powers of z since coshz is an even function; that is, E2tt+l
=0
(n = 0, 1, 2, . ..). To find the series, we divide the series
coshz = 1+—+— + —+•••= l +-z2 +—z4+—z6+"- (lzl< °°)
2! 4! 6! 2 24 720
into 1. The result is
,1 2 5 4 61 6 (i i^ n>
\= 1
—
z +— z z +••Ilzl<— I,
coshz 2 24 720 V 2
or
93
1 , 1 2 5 4 61 6 f. . n= 1 z +—z z +•••
I lzl<—coshz 2! 4! 6! V 2
Since
_!_ = £ +^ + Jlz4 + JSlz«+... f|
z|<f|,coshz 2! 4! 6! V 2
this tells us that
£0=1, E2= -1, £4
= 5, and E6=-61.
94
Chapter 6
SECTION 64
1. (a) Let us write
_i- = I.-i- = I(i- z + z2-z
3 + ...) = i-l + z -z2 +... (0<lzkl).
z + z z 1 + z z z
The residue at z = 0, which is the coefficient of -, is clearly 1.
z
(b) We may use the expansion
cosz = l-2! 4! 6!
to write
m 1 1 1 1 1 1 "\ 111111ZCosH = z^l--.? + -.7
--.7 +
...J
= z--.- +-.-3---7 -f.
(0<lzl<°°).
The residue at z = 0, or coefficient of -, is now seen to be -—
.
z 2
(c) Observe that
z-sinz 1. . . 1= -(z-smz) = -
z z z r 3! 5!*'j
T2 T4
3! 5!
(0<lzl<<*>).
Since the coefficient of - in this Laurent series is 0, the residue at z = is 0.
z
(d) Write
and recall that
cotz _ 1 cosz4 ~ 4 '
.
z z sinz
T2 T4 T2 -T4
1 z,z , z z
cosz = 1 + = 1 +2! 4! 2 24
(lzl< ~)
and
Dividing the series for sinz into the one for cosz, we find that
95
cosz 1 z z3
Thus
cotz _ I fl z 11 11z4 "z4 U 3 T5
+'-)-7~37~V~z
+- (0 <&<*).
Note that the condition of validity for this series is due to the fact that sinz = when
z = nit(n = 0,±l,±2,...). It is now evident that has residue —-. at z = 0.
(e) Recall that
z4
45
sinhz = z +|y
+^ + ...(| z |
<oo)
and
1-Z
There is a Laurent series for the function
1i 2= l + z + z
2 + -.- (lzl<~).
sinhz 1 . ( 1 ^
that is valid for <lzl< 1. To find it, we first multiply the Maclaurin series for sinhz
and1-z
2
,
1 3 1 5z + -z:
+ z +•6 120
3 1 5z +-z +•
z5 +-
7 3z + tz +••• (0<lzl<l).
o
96
We then see that
sinhz 17 1=— + -•-+ ••• (0<lzl<l).
z4(l-z
2
) z3
6 z
This shows that the residue of 47—^ at z = is —
.
z (1 — zJ
6
2. In each part, C denotes the positively oriented circle lzl= 3.
(a) To evaluateJc—^
—
dz, we need the residue of the integrand at z = 0. From
the Laurent series
exp(-z) _ 1
2—
_2
^ z,z2
z3
, ) 1 1 1 1 z ,A . .,1— + +...=_ + +... (0<lzl<°°),
^ 1! 2! 3! J z2
1! z 2! 3!
we see that the required residue is -1. Thus
lc™P^
dz = 2n i-l)= -2jti .
(c) Likewise, to evaluate the integraljc
z2 exp^jdz, we must find the residue of the
integrand at z = 0. The Laurent series
, • , 11 1 1 11 11
2 Z 1 1 1 1 1= Z +— +— + +
1! 2! 3! z 4! z2
which is valid for <lzl< °°, tells us that the needed residue is — . Hence6
97
r z + 1
(d) As for the integral -=
—
—dz, we need the two residues ofJC 7* — 9rz l
-2z
z+1 z + 1
z2-2z z(z-2)'
one at z = and one at z = 2. The residue at z = can be found by writing
z + 1
z(z-2) z-2) { 2A z) l-(z/2)
\ 2 2 z)
f z z1 %
2 22
which is valid when 0<lzl<2, and observing that the coefficient of - in this last
z
product is — . To obtain the residue at z = 2, we write
z + 1 (z-2) + 3 1
z(z-2) z-2 2 + (z-2) 2V z-2j l + (z- 2)/2
2\ z-21
z-2|
(z-2)2
which is valid when <lz - 21< 2, and note that the coefficient of in this productz-2
3is — . Finally, then, by the residue theorem,
Jc z2-2z V 2 2)
In each part of this problem, C is the positively oriented circle lzl= 2.
(a) If/(z)l-z :
-, then
/ - =-—4=-—-;
—
r = -- l+z3+z
6+••• =--—
z
2-\zj z -z z 1-z z
v yz z
1_2
when <lzl< 1. This tells us that
f(z)dz = 2mRes\f(-) = 2ni(-l) = -2m.<=° z \z)
(b) When f(z) =1
2 > we have1 + z
Thus
f = 2mRes 4"/f- ] = 2m(0) = 0.
(c) If /(z) = -, it follows that ~/[-) = -• Evidently, then,Z 2 VZ/ Z
f f{z)dz = 2m Res -^/f- ] = 2/ri(l) = 2m.ic z=o z
z\ Z J
Let C denote the circle lzl= 1, taken counterclockwise.
(a) The Maclaurin series ez =T— (lzl< °°) enables us to write
(b) Referring to the Maclaurin series for ezonce again, let us write
Now the - in this series occurs when n-k = -l, or k = n + l. So, by the residuez
theorem,
l> expf-ljz =2m-i— (/j = 0,1,2,...).Jc vz/ (« + l)!
The final result in part (a) thus reduces to
995. We are given two polynomials
P(z) = a +alz + a2z
2+--- + a„z'' (a„*0)
and
<2(z) = b + blZ + b2z2 + --- + bmz
m(bm * 0),
where m>.n + 2.
It is straightforward to show that
J_ pQ 1 z> - flQz'""2 + a
iz"" 3 + a2*
m~4+ • • • + fl^
w"""2
z2 "
2(1 / z) V" + fri*"*"1
+V"2+ • • • +K
1
Observe that the numerator here is, in fact, a polynomial since m -n - 2 > 0. Also, since
bm * 0, the quotient of these polynomials is represented by a series of the form
do+diZ + dzZ2 + ••-. That is,
1 P(l/z) j , j 2(0<lzl</^);
L 1 P(\lz) . ^ rtand we see that —r has residue z = 0.
z 2(1 /z)
Suppose now that all of the zeros of Q(z) lie inside a simple closed contour C, and
assume that C is positively oriented. Since P(z)/Q(z) is analytic everywhere in the finite
plane except at the zeros of <2(z), it follows from the theorem in Sec. 64 and the residue just
obtained that
P(z)dz = 2mRes
z=0
J_ P(l/z)
z2'ea/z)
= 2^*0 = 0.,c Q(z)
If C is negatively oriented, this result is still true since then
Piz) J__ f P(z)
icQ{z) J-c Q(z)
dz = 0.
SECTION 65
1. (a) From the expansion
we see that
1! 2! 3!
,n ^,^11 iizexp| _j =z+1+_._ +_._ + ..
(Izl<~),
(0<lzl<oo).
100
The principal part of zexpl - I at the isolated singular point z = is, then,z
2!'z+3!Y
+ "';
and z = is an essential singular point of that function.
(b) The isolated singular point of is at z = -1. Since the principal part at z = -11 + z
involves powers of z + 1, we begin by observing that
z2 = (2 + 1)
2 - 2z - 1 = (z + 1)2 - 2(z + 1) + 1.
This enables us to write
e _ (z+lf -% + !) + ! 1
1+z z+1 z+1
Since the principal part is —^— , the point z = -1 is a (simple) pole.,
z + 1
(cj The point z = is the isolated singular point of £m-£
) and we can write
z
sinz 1 ^-— +— -•••1 = 1-— +— -••• (0<lzl<~).z z\, 3! 5! J 3! 5!
The principal part here is evidently 0, and so z = is a removable singular point of the
-t
. smzfunction .
cos z(d) The isolated singular point of is z = 0. Since
z
cosz 1 ( z2
z4
I 2! 4! ;
=I_±+ £l_... (0<lzl<oo),
z 2! 4!
I COSZthe principal part is — . This means that z = is a (simple) pole of .
z z
(e) Upon writing —^—-j = we find that the principal part of3
at its
(2 — z) (z-2) (2-z)
isolated singular point z = 2 is simply the function itself. That point is evidently a pole
(of order 3).
(a) The singular point is z = 0. Since
101
1-coshz 11-
( _2 ,4 6
!+£.+£.+£_+..V 2! 4! 6! J 2\ z 4! 6!
when <lzl< °°, we have m = l and B =—— = -—
.
2! 2
(ft) Here the singular point is also z = 0. Since
l-exp(2z)_ 1 ( 2z 2V 2 373 2V 2V "\
I 1! 2! 3! 4! 5!
= _2_ J__2^ _1 2j_ _j__2^_2^ll'z
3
2!*z2
3!'z 4! IT*"
23
4when <lzl< °°, we have m = 3 and B = =—
.
3! 3
(c) The singular point ofexP^z
^ js z = \ jhe Taylor series(z-1)
exp(2z) = e2(z-V=e 1 +
2(z-l)l22(z-l)
2
.2
3(z-l)
3
1! 2! 3!
(Izl< oo)
enables us to write the Laurent series
exp(2z) _ ^(z-1)
2
2+— •
1 22
22
+— + -J-(z-l)+-L(z-lr 1! z-1 2! 3!
(0<lz-ll<°°).
2Thus m = 2 and = e
2— = 2e2
.
Since / is analytic at Zq, it has a Taylor series representation
f(z) = f(z )+^-(z-z ) +^^-(z-z )
2 +- (Iz-ZoK^,).
Let g be defined by means of the equation
102
(a) Suppose that /(z ) *0. Then
g(z) =1
/<0 +^(*-*) +^U-*o)2 +-
- - 1! 2!° ;
Z-Zn(0<k-z l</?o).
This shows that £ has a simple pole at Zq , with residue /(z ).
(b) Suppose, on the other hand, that /(z ) = 0. Then
1! 2!
(0<\z-z \<Ro).
Since the principal part of g at z is just 0, the point z = is a removable singular
point of g.
4. Write the function3_2
/(Z) =(?W (a>0)
as
_ 4Kz)3_2
8a3z
/(z) =/ -n3
where 0(z) = v .n3-
(z + «T
Since the only singularity of 0(z) is at z = -ai, 0(z) has a Taylor series representation
Hz) = ««0 + *M(Z - ai) +^(z- ai? + (lz-ai'l<2a)
about z = ai. Thus
/(z) =1
(z-m) 30(ai) + ^rr^Cz - fl + (z - ai) +
1! 2!(0<lz-ail<2a).
Now straightforward differentiation reveals that
16a4/z-8fl
3z2
. , 16a3(z
2-4fl/z-a
2)
^ (z) =—; , „ 4
— and <j> (z) = -(z + ai)
4(z + ai)
3
Consequently,
Q(ai) = -a2i, 0'(a/) =™, and </>"(ai)--i.
This enables us to write
/(Z) =(z-aif [~
a2i ~f(Z ~ fll) ~i
(z " m)2 + "']
(0 < lz - ml < 2a).
The principal part of/at the point z = ai is, then,
ill all a2i
z - ai (z - ai)2
(z - ai)3 '
SECTION 67
1. (a) The function f(z) -Z + 2
has an isolated singular point at z = 1. Writing /(z) =z~l z-1
where (/>(z) = z2+2, and observing that 0(z) is analytic and nonzero at z = 1, we see
that z = 1 is a pole of order m = 1 and that the residue there is 5 = 0(1) = 3.
(b) If we write
f(z) =lz + 1
Hz)
z- -zrf, where 0(z) =
1
8'
we see that z = -- is a singular point of /. Since 0(z) is analytic and nonzero at that
point, /has a pole of order m = 3 there. The residue is
B _ r(-V2) _ 3
2! 16'
fc) The function
expz expz
z2 + ;r
2(z-/n)(z + 7ri)
has poles of order m = 1 at the two points z = ±m. The residue at z = to is
Bi
exp TO _ -1 _ i
2 to 2 to~ In
'
and the one at z = -ni is
1041/4
2. (a) Write the function /(z) = (I zl> 0, < arez < 2 it) asz + 1
Hz)/(z) = z^7. where <p{z) = z
m =e 4 (lzl>0,0<argz<2;r).z + 1
The function 0(z) is analytic throughout its domain of definition, indicated in the
figure below.
Also,
-1 o
• Branch cut
jl / n , ni/4 _ J'osH) Ja«i+<*) iwM it . . it 1+i<9(-l) = (-1) =e 4 = £ 4 = e = cos— + zsin— = —f=-*0.
4 4 V2
This shows that the function /has a pole of order m = 1 at z = -1, the residue there
being
5 = 0(-l) =1 + /
V2'
(b) Write the function /(z) =Logz
(z2+ D
2as
From this, it is clear that /(z) has a pole of order m = 2 at z = i. Straightforward
differentiation then reveals that
Logz tf + 2iRes & = =
(z2 + l)
28
(c) Write the function
.1/2
/(*) =(z
2+ D
2
105
(ld>0,0<argz<2;r)
as
Since
and
f(z) =.1/2
(z-i)2
where =(z + 2
f W2(z + /)
3
r1/2 - *-''*/4 - _L _i_ ,-i/2 _ j*/4 _ 1.
'
71/2
1 — '
Res f „„ = ft'(t>1
*=<~(z2+ l)
2
(a) We wish to evaluate the integral
8V2"
3z3 + 2
,c (z-l)(z2+9)
dz,
where C is the circle \z - 21 = 2 , taken in the counterclockwise direction. That circle and
the singularities z = 1, ± 3i of the integrand are shown in the figure just below.
Observe that the point z = 1, which is the only singularity inside C, is a simple pole ofthe integrand and that
Res-3z
3 + 2 3z3 + 2
(z-l)(z2+9) z
2 + 9
According to the residue theorem, then,
3z3 + 2
1
2
liz-\Z^9)
dz = 1M{l)
=ni -
106
(b) Let us redo part (a)when C is changed to be the positively oriented circle Izl = 4, shownin the figure below.
In this case, all three singularities z = 1, ± 3/ of the integrand are interior to C. Wealready know from part (a) that
3z3 + 2 1
Res-*=1 (z-l)(z
z+9) 2
It is, moreover, straightforward to show that
3z3 + 2 3z
3 + 2Res -
<= 3i (z-l)(z' + 9) (z-l)(z + 3/).
and
15 + 49/
12
Res-3z
3 + 2 3z3 + 2
z=-3,-(z_ +9) (z _ i)(z _ 3/)
The residue theorem now tells us that
j=-3i
15-49/
12
f 3z3+ 2 , _ /l 15 + 49/ 15-49/"\ ,5 dz = 2m\-+ + =6
}c(z -l)(z2+9)
"V2 12
m.
4. (a) Let C denote the positively oriented circle Izl = 2, and note that the integrand of the
r dzintegral -r- — has singularities at z = and z = - 4. (See the figure below.)
Jc z (z + 4)
y
To find the residue of the integrand at z = 0, we recall the expansion
107
and write
(lzl<l)
1 _ 1 1
z\z + 4) 4z3
_l + (z/4)
_n-3
Now the coefficient of - here occurs when n = 2, and we see thatz
1 1
Consequendy,
Res ,*=° z\z + 4) 64
Jc z\z + 4) \64) 32
(0<lzl<4).
(b) Let us replace the path C in part (a) by the positively oriented circle \z + 21 = 3, centered
at -2 and with radius 3. It is shown below.
We already know from part (a) that
ReS^ —
.
*=° z\z + 4) 64
To find the residue at -4, we write
1 _ <t>(z) . ... 1
Tw-T+Ty where Hz)= 7-
This tells us that z = -4 is a simple pole of the integrand and that the residue there is
0(-4) = -1 / 64. Consequendy,
J<^ 5(z+4) {64 64J
108
f cosh 7tz dzLet us evaluate the integral :—7T"» wnere C is the positively oriented circle lzl= 2.
JC TIT * 1 1^ Z^Z + ij
All three isolated singularities z = 0,±i of the integrand are interior to C. The desired
residues are
_ cosh nz cosh nzRes—s =—5
z=o z(r+l) z +1 .
=1,
_ cosh 7cz cosh tfzRes—= =
z(z2+ l) z(z + i)_
1
.~2'
and
_ cosh nz cosh tezRes— =*=-<"z(z
2+l) z(z-i)
_ 1
.~2'z=-i
Consequendy,
r coshflzdz „ /, 1 1, . .
= = 2m 1+—+— \= 4m.
,c z(z2+l) V 2 2,
6. In each part of this problem, C denotes the positively oriented circle lzl= 3.
(a) It is straightforward to show that
f„\ Oz + 2)2
1 /n_ (3 + 2z)2
lf/(z) =z(z-l)(2z + 5)'
111611
ZllJ-zd-^ + Sz)-
This function -y/^-j has a simple pole at z = 0, and
I(3z + 2)2— dz = 2/ri Res
cz(z-l)(2z + 5) *=° z/J= 2/ri
<2,= 9ni.
(b) Likewise,
if/(z)=z\l-3z)
±en*-3
'w(l+z)(l + 2z
4)
-*-M-lz2 U; z(z + l)(z
4+2)
The function \ )has a simple pole at z = 0, and we find here that
f
Z(1 " 3z)
4<fe = 2;riRes
Jc(i + z)(l + 2z4)
*=°V 1
= 27ri| -- |= -3;ri.
(c) Finally,
109
zV"then ±f^=7f+
1 Jl
z3
)
The point z = is a pole of order 2 of -jf -. The residue is 0'(O), where
z \z
<p(z) = -
Since
<t>\z)
1 + z3
'
(l + z3)e
z -e l3z
2
3\Z
the value of 0'(O) is 1. So
f'c 1 + z
3dz = 2m Res 7f
il)=2jtiW =2m -
SECTION 69
1. (a) Write
esc z = = -^4, where /?(z) = 1 and q(z) = sin z.sinz q(z)
Since
p(0) = l*0, ^(0) = sin = 0, and tf'(0)= cos0 = l*0,
z - must be a simple pole of cscz, with residue
P(0) =- = 1q'(0) 1
(b) From Exercise 2, Sec. 61, we know that
^ 1. 1 1
CSCZ = - +—Z+ Tz 3! L(30
25!.
1
z3+- (0 <lzl< it).
Since the coefficient of - here is 1, it follows that z = is a simple pole of cscz, the
residue being 1.
(a) Write
z- sinhz p(z)
z2sinhz
=^)' 6 ^) = *- sinh * and tf(z) = z
2sinhz.
Since
p(Ki) = m*0, q(m) = 0, and (m) = j? * 0,
it follows that
Rc „ z-sinhz = p(ni)= m__i_
*=*«' z2sinhz q'(,ni) n1 k
(b) Write
exp(zr) p(z)
sinhz=
~q(zj'6 ^)
= exP(*') ?(z) = sinhz.
It is easy to see that
«-*» sinhz q\7ti)r
*=-*/ sjnhz g'(-»0P
Evidently, then,
Resexp(£ + Res
expU0 = _ 2 «Pff*) + CTpW*)sinhz sinhz 2
(a> Write
Observe that
/(z) = ^7T' where P(z) = z and o(z) = cosz.
«(f + n«:j = (n = 0,±l,±2,...)..2
Also, for the stated values of n,
p\- + n7c\ =-+ n7t* and ^| +«^ = -sin^| +n^ = (-l)fl+1
^0.
Ill
So the function f(z) = has poles of order m = 1 at each of the pointscosz r
~"2
z„=-+nn (n = 0,±l,±2,...).
The corresponding residues are
(b) Write
tanh ^ = where p(z) = sinhz and q(z) = coshz.
Bothp and q are entire, and the zeros of q are (Sec. 34)
z = I ^- + nw \i
2nnf (n = 0,±l,±2,...)
In addition to the fact that J (~ + jfJ
= 0, we see that
Pi[j+ nn
)*)= sinh(^ + = icosnn = /(-!)" *
and
*'((f+ = Sinh
(f '
+ n?n)=
* °-
So the points z = + (« = 0,±1,±2,...) are poles of order m = 1 of tanhz, the
residue in each case being
p\ [— + nn }i I
Let C be the positively oriented circle lzl= 2, shown just below.
TO 72 x
(a) To evaluate the integralJc
tan z dz , we write the integrand as
tanz = £^£l) where p(z) = sin z and q(z) = cosz,
and recall that the zeros of cos z are z = -^ + n;r (n = 0,±1,±2,...). Only two of those
zeros, namely z = ±tt/2, are interior to C, and they are the isolated singularities of
tanz interior to C. Observe that
n(nll)Res tanz = ,
' =-1 and Res tanz • „ _z=t/2 tf(tf/2) qX-npi)
_ p(-x/2) _ L
Hence
f tanz*fe = 27ri(-l-l) = -47ri.
(fe) The problem here is to evaluate the integralJ
integrand as
c sinh2z. To do this, we write the
—\— =£Qt where p(z) = l and g(z) = sinh2z.
sinh2z q(z)
Now sinh2z = when 2z = nm (n = 0,±1,±2,...), or when
— (/z = 0,±l,±2,...).
2
Three of these zeros of sinh2z, namely and±-^-> are inside C and are the isolated
singularities of the integrand that need to be considered here. It is straightforward to
show that
Res_J_ =£W =_J_ = I*=o sinh2z q'(0) 2cosh0 2'
Res
113
1 p(m/2) _ 1 1 1
*=»/2sinh2z q'(m/2) 2cosh(;ri) 2cosrc 2'
and
Res1 = PtE!2± = 1
= 1 = *
z=-«v2 Sinh2z q'(-jti/2) 2cosh(-;rz) 2cos(-;r) 2'
Thus
Jc sinh2z U 2 2-to.
5. The simple closed contour CN is as shown in the figure below.
Within Q, the function -= has isolated singularities at
z smz
z = and z = ±nn (n = l,2,...,AT).
To find the residue at z = 0, we recall the Laurent series for cscz that was found in Exercise
2, Sec. 61, and write
1 1 1 1 1= -t-cscz = -ri - +—z +
z sinz z2
z2\z 3!
1 1
(3!)z
5!z3+-
1 1 1-=+ +z
36 z
1 1
(3!)2
5!_z+- (0<IzI<tt).
114
This tells us that -=4— has a pole of order 3 at z = and thatz smz
1 1Res«=° z
i smz 6
As for the points z = ±nn (n = 1,2, . . . , N), write
~Y~— = ^tt, where p(z) = 1 and q(z) = z2sinz.
z smz q(z)
Since
p(±rut) = 1*0, q(±nn) - 0, and q\±nn) = n2n2
cosnit = (-l)"nz7F
z56 0,
it follows that
1 1 (-1)" (-1)"Res*=±»* z
2sinz (-1)"«V (-1)"
~~ nV
So, by the residue theorem,
J,c" z sinzcfe = 2m
6
Rewriting this equation in the form
A (-1)"+1
= tt2
g f dz
il n2
12 4i'-kz2sinz
and recalling from Exercise 7, Sec. 41, that the value of the integral here tends to zero as Ntends to infinity, we arrive at the desired summation formula:
f (-ir1
= n2
it n2
12
The path C here is the positively oriented boundary of the rectangle with vertices at the
points ±2 and ±2 + i. The problem is to evaluate the integral
dz
c(z
2-l)
2 +3'
The isolated singularities of the integrand are the zeros of the polynomial
115
q(z) = (z2-l)
2+3.
Setting this polynomial equal to zero and solving for z2
, we find that any zero z of q(z) has
the property z2 = l±V3i. It is straightforward to find the two square roots of 1+ V3i and
also the two square roots of 1-V3i. These are the four zeros of q(z). Only two of those
zeros,
z,.^-.^*L and d*+L,
he inside C. They are shown in the figure below.
To find the residues at z and -Zq , we write the integrand of the integral to be evaluated as
1 _ P(z)
(z2-l)
2 +3 q(z), where p(z) = 1 and q(z) = (z -
1) +3.
This polynomial q(z) is, of course, the same q(z) as above; hence q(z ) = 0. Note, too, that
p and q are analytic at Zq and that p(z ) * 0. Finally, it is straightforward to show that
q\z) = 4z(z2 - 1) and hence that
?'(*<,) = 4z (4 -1) = -2V6 + 6V2/ * 0.
We may conclude, then, that z is a simple pole of the integrand, with residue
P(z ) _ 1
q'(z ) -2V6+6V2i'
Similar results are to be found at the singular point -Iq. To be specific, it is easy to see that
qX-Zo ) = -q'(z„ ) = -7(^) = 2V6 + 6V2i * 0,
the residue of the integrand at -Zq being
q'i-Zo) 2V6 + 6V2/*
Finally, by the residue theorem,
f = 2ml1 1 __
^
(
Z2 -
1)2 + 3 \-2sf6 + 6V2i
+2V6 + 6V2i J
=2^2
We are given that /(z) = l/[?(z)]2
, where ? is analytic at z*, ?(z ) = 0, and ?'(z )*0.These conditions on 4 tell us that q has a zero of order m = l at v Hence#(z) = (z - Zq)^(z), where £ is a function that is analytic and nonzero at z ; and this enablesus to write
f(z) =-^-T , where =—^.
So /has a pole of order 2 at z , and
Res/(z) = 0'(zo )=-M^.
But, since q(z) = (z -z )#(z), we know that
q'(z) = (z-z )g'(z) + g(z) and ?"(z) = (z-z )g,,
(z) + 2g'(z ).
Then, by setting z = z in these last two equations, we find that
q'(z ) = g(z ) and ?"(z ) = 2s'(z ).
Consequently, our expression for the residue of/at zQ can be put in the desired form:
Res/(z) = - q"^\.
(a) To find the residue of the function esc2z at z = 0, we write
csc z ~r < sn2 »
wnere ^(z) = sinz.
W(z)J
Since q is entire, #(0) = 0, and q'(0) = 1 * 0, the result in Exercise 7 tells us that
Rescsc2z = - q = 0.
[<z'(0)]3
(b) The residue of the function -
—
^-—j at z = can be obtained by writing(z + z
)
1
-, where q(z) = z + z2
{z + z2
) [q(z)f
Inasmuch asqis entire, q(0)=0, and q'(0) = 1*0, we know from Exercise 7 that
Res 1 =-j£<°>_ = -2 .- (z+z1
) [«W
118
Chapter 7
SECTION 72
1. To evaluate the integral J——
-, we integrate the function f(z) = around the simplen x +1 Z +1
closed contour shown below, where R>L
y
c.
We see that
where
_Jsx
2 + l J^ z2 +l
Thus
5 = Res —r—— = Res«' z +1 (z-f)(z + Z + l'_U 2i
f f <fe
Now if z is a point on Q,
lz2 + ll>llzl
2-ll=rt
2-l;
and so
J,
E.
R2 -l i_J_R2
-»0 as
Finally, then
1 dx 1 dx n1—7 = tt> or —: =—
.
_J *
2+ 1 Jjc2 +1 2
119
°° dx 1
2. The integralJ—5—rrj" can be evaluated using the function f(z) =— 5- and the sameJ (x +1) (z + 1)
simple closed contour as in Exercise 1. Here
_((jc2 + l)
2+fc(z
2 + i).2 , i\2
= 27riB,
where B = Res1
*"<~V + 1)2
. Since
1 Hz) . . , x1
12—^ = TT-^' where 0(z) =—
-
U +(z2 + l)
2(z-i)
2
we readily find that B = <j>'(i) = -7- , and so4i
J„0t2 + 1)
22 J<
fife
?(*
2 + l)2
2 ^(z2+l)
2
If z is a point on CR , we know from Exercise 1 that
lz2 + ll>i?
2-l;
thus
dzt az
b (z2 +(z2 + l)
2
The desired result is, then,
nR J? 3<—i t= , ,2 ~>0 as R-*°°.
°r dx _% r dx %_
i(x2+i)
2 "2' orJcTTif'T
3. We begin the evaluation of[ f* by finding the zeros of the polynomial z* + 1, which are
+1
the fourth roots of -1, and noting that two of them are below the real axis. In fact, if we
consider the simple closed contour shown below, where R>1, that contour encloses only
the two roots
jit/* _ 1 1
k ~ €"V2
+V2
and
J3X/4 iJT/4„iJt/2 1 . / V 1 i
120
Now
where
R - Res—^— and 2?, = Res—
r
*=*2 Z + 1
The method of Theorem 2 in Sec. 69 tells us that z, and z1 are simple poles of —:
andz +1
that
^=rr-- = -7- 311(1 fi2 =TT-- = -T'4zj Zi 4 4z2
3z2 4
since z,4 = -1 and zt = -1. Furthermore,
Bl+B
2 =-j(zl+z2 )
=~2V2"
Hence
Since
t dx _ n t dz
}R x* + l~j2 Jc, z* + i'
J,c«z4 +l
we have
°°t dx n °r dx n
iy + l"V2'°r L4 + l"2V2-
121
r x2dx
We wish to evaluate the integral —=———=—— . We use the simple closed contourl(x +l)(x
2+4)
shown below, where R>2.
We must find the residues of the function /(z) =—, : at its simple poles(z + l)(z + 4)
z = i and z = 2i. They are
Bl= Res/(z) =
(z + /)(z2 + 4).
and
Thus
*=2< (z2+l)(z + 2z).
fi2=Res/(z) =
*=2<(7
Z
?=2i
6/
3/'
f*A +f
K(x2+l)(x
2 + 4) Jc
z2dz
or
ijtf + l)(x' + 4) Jc. + + 4)
* x2dx n r z
2dz
=--f —JR(x
2+l)(x
2+4) 3 Jc(z3 + lXz
2 + 4)
If z is a point on CR , then
lz2 + ll^llzl
2-ll = /?
2 -l and lz2+4l2>lld
2-4l = /?
2 -4.
Consequendy,
z2<fc
(z2 + l)(z
2+4)
and we may conclude that
(R2-l)(R2 -4)
-»0 as
x dx n
i(jf" +l)(^ + 4) 3 J(jt
2 + l)(*2 + 4) 6
dx it
122
f x dx5. The integral f—5———5——5- can be evaluated with the aid of the function
{(x2 + 9)(x2 + 4)
2
f(z)(z
2+9)(z
2+4)
2
and the simple closed contour shown below, where R > 3.
y
We start by writing
J
x dx
R{x
2 + 9){x2 + 4)
2 Jct(Z
2 + 9)(^ + 4)2
z2dz
where
Now
z2
z2
A = ResTi—r~n— and 5, = Res—5 = =-.
«-* (zz + 9)(z
z+4)
z
t1
(z + 3i)(z2 + 4)
2
J?=3<
(zz+9)(z
z
+4r
3
50/'
To find B2 , we write
z2
__0(z) ._, x , , Z"
(z2+9)(z
2 + 4)2
(z-2/):
-, where 0(z) =
Then(z
2+ 9)(z + 2/)
2
2?2 = 0'(2/) =
This tells us that
13
200/
'
Ax
2dx
JR(x
2+9)(x
2+4)
2100 Mza + 9)(z
a+4)
:
=JL_f100 J<Wz2 +
z2</z
.2 . a\2'
Finally, since
we find that
1,
z2dz
c*(z2+9)(z
2+4)
2
7r/?3
(R2 -9)(R2 -4)
j -» as R -> 00,
f 0c2 +
*2d!x 7T
(*z+9)(;c
2 + 4)2
100or
1
* etc
U2+9)(a:
2+4)2 200'
7. In order to show that
P V f
xdx= -—
'^(x2 + l)(x2 + 2x + 2) 5'
we introduce the function
(z2+ l)(z
2+2z + 2)
and the simple closed contour shown below.
y
/ x >*
.
R X
Observe that the singularities of /(z) are at i, z„=-l + i and their conjugates
z = -1 - i in the lower half plane. Also, if R > V2, we see that
K
jf(x)dx + j f(z)dz = 2m(B +Bl ),
where
and
2? =Res/(z)=*
«-«o (zz+l)(z-z )
Bl=Resf(z) =
Z=Zo
= _JL _L-io
+io
I
Evidently, then,
(z + /)(z +2z + 2)
f £<& = _£_f zrfz
4U2 + 1)(jc2 + 2x + 2) 5 ^(z2 + l)(z
2 + 2z +
= _L_I-10 5
1 '
zdz
Since
r Z*fe r
h (z2 + l)(z
2 + 2z + 2)~
(z2 +(zz + l)(z
i + 2z + 2)
as R -» °°, this means that
zdz
(zz+ l)(z-z )(z-z^)
,. r xdx nhm —; =—
.
*-*~L(x2+l)(x
2 + 2x + 2) 5
2)
(tf2-!)(/? -V2)
2->0
This is the desired result.
124
8. The problem here is to establish the integration formulaJ— - = using the simple
closed contour shown below, where R > 1.
1
There is only one singularity of the function f(z) =
to the closed contour when R > 1 . According to the residue theorem,
dz . r dz . r dz
3 , namely z - eiicn
, that is interior
z +1
JclZ3 +l Jc« z
3 + 1 JC2Z3 + 1 z=Zo Z
3 +l
where the legs of the closed contour are as indicated in the figure. Since Q has parametric
representation z = r (0 < r < R),
fz = f
dr
kz 3 + l~{ r3 + l'
and, since -C2 can be represented by z = reann
(0 < r < R),
dzJ- *
J2*nf dz r dz _ f e
,Kdr _ n,/3 f
JcJZ3 + i J-c2Z 3 + 1 J(re ,2,c/3
)3 + l Jr3 + 1'
Furthermore,
Consequently,
But
5S.V + 1 3z2- ~ 3*
i2 *' 3
U V + l 3^2ff/3 Jca2> +r
•h z3 + i
. 1 2^i? n< > as R-><x>.R3 -l 3
This gives us the desired result, with the variable of integration r instead of x :
fdr 2
JV + 1 3(ea*n -e l
2m 2ni n 2%
» t A 3(e -e -e ) 3(e —e ) 3sm(2^/3) 3\3
125
Let m and n be integers, where < m < n. The problem here is to derive the integrationformula
[—5-
—
-dx = —esci x
2n + 1 In \ In
(a) The zeros of the polynomial z + 1 occur when z2n - -1. Since
.(2k + i)n
In(-l)
1/(2B) =exp
it is clear that the zeros of z2n + 1 in the upper half plane are
and that there are none on the real axis.
(b) With the aid of Theorem 2 in Sec. 69, we find that
(k = 0,1,2,.. .,2n-l),
(* = 0,l,2,...,n-l)
2m 2m ,
Res-^— = .
c\_ ,
= J_c*(—)+>
<-»V + l 2wc,z''- 1
2n2n-l (* = 0,1,2... .,n-l).
^ . 2m + lPutting a = n, we can write
In
2(m-n)+l |. (2fc + l)7r(2m- 2w + i)l" eXT In J
[\(2* + l)(2m + l);rl , , N.
= expl i-i ^- '— exp[-/(2fc + l);r] =
Thus
-2m
Res-*=<*z
2" + l In
(2t+l)o(* = 0,1,2 /i-l).
In view of the identity (see Exercise 10, Sec. 7)
n-l i n
k=0 1 ~
126
then,
n-l ,2m n-l
f^z2n + l
m— el-ei2an
e-ia
n l-e:2a -ia
m ei2an
-l
n'
eia-e~
ia
m eiitm+l)*-l n 2i n
n e -e n e'a -e ,a nsma
(c) Consider the path shown below, where R>1.
y
c.
The residue theorem tells us that
r x2mr z
2m54 r2m
J T57TT^ +L 3TTT <fe = 2;R'2 Res^T'Zr x +1 JC,, Z +1 k=o
z=c"Z +1
or
}R x2n +l nsina k z
2n +l
Observe that if z is a point on CR , then
lz2m
l = /?2m
and \z2n + \\>R2n
-I.
Consequently,
W'+l dzR2m
R2a -l "/?
-2»1
-2n
2(n-m)-l
1-1
->0;
2«
and the desired integration formula follows.
10. The problem here is to evaluate the integral
Ji
dx
Kx2-a) 2 + l]
2 '
where a is any real number. We do this by following the steps below.
127
(a) Let us first find the four zeros of the polynomial
q(z) = (z2-a)2 + l.
Solving the equation q(z) = for z2
, we obtain z2 =a±i. Thus two of the zeros are
the square roots of a + i, and the other two are the square roots of a - i. By Exercise
5, Sec. 9, the two square roots of a + i are the numbers
Zq =-j=>(^A + a+rtA-a) and -z ,
where A = 4c? + 1. Since (±z )
2=z\ =a + i = a- i, the two square roots of a - i, are
evidently
z and -z .
The four zeros of q(z) just obtained are located in the plane in the figure below, which
tells us that z and -Zo lie above the real axis and that the other two zeros lie below it.
•
y
•
Zo
X
• •
-Zo Zo
(b) Let q(z) denote the polynomial in part (a) ; and define the function
1
f(z) =[q(z)f
which becomes the integrand in the integral to be evaluated when z = x. The method
developed in Exercise 7, Sec. 69, reveals that z is a pole of order 2 of /. To be
specific, we note that q is entire and recall from part (a) that q(Zo) = 0. Furthermore,
q\z) = 4z(z2 - a) and z\=a + i, as pointed out above in part (a). Consequently,
2'(z ) = 4z„ (Zq -a) = 4iz * 0. The exercise just mentioned, together with the relations
zl=a + i and 1 + a1 = A2
, also enables us to write the residue of/at Zq'.
g"(Zo) = 12z2 - 4a _ 3zp
2 - a _ 3(a + i)-a a — i __ a — /(2a2+ 3)
WW (4iz )3
16iz2z 16i(a + i)zo' a-i 16A
2Zo
As for the point -z , we observe that
q'(-z) = -q\z) and q'\-z) = q'Xz).
Since q(-z ) = and #'(-z ) = -q'do) = 4i'z 0, the point -Zq is also a pole of order
2 off. Moreover, if B2denotes the residue there,
_ <?"(-?<)) _ <?"(*(,) _ f g»frb) 1 _[^'(-^,)]
3
[<?'(z )]
3
l^'(^o)]3
= -5x
.
Thus
B.+B2= B.-B.= 2r'Im5. =—Viml 2 1 1 1
8A2/
-g + f(2fl2+3)
Zn
We now integrate /(z) around the simple closed path in the figure below, where
R >\z \ and CR denotes the semicircular portion of the path. The residue theorem tells
us that
R
jf(x)dx + f(z)dz = 27ri(A + B2 ),
or
I;
dx
In order to show that
Im-a + i(2a
2+3) dz
[<?(z)J2
•
lim f
dx= 0,
*^~ Jc« [?(z)]
2
we start with the observation that the polynomial q(z) can be factored into the form
q(z) = (z - Zq )(z + Zo )(z ~ Zo )(z + z )•
lz±Zol>llzl-IZoll = /?-lzol and \z±Zq\>\ Izl-lz^l I = R-\z \.
This enables us to see that \q(z)\ ^ (/?-lz„l)4when z is on CR . Thus
129
1
(R-\z \)'
for such points, and we arrive at the inequality
nR
(tf-IZoO8 8 '
which tells us that the value of this integral does, indeed, tend to as R tends to <».
Consequently,
P.V.j
dx n
_J.[(*
2 -a)2+l]
2 4A2
But the integrand here is even, and
Im-q + /(2a
2 + 3)
^0
Im-a + i(2a
z+3)
Zo
= Im V2-a + i(2a
2 + 3) -jA + a - i-yjA-a
VA + a +/VA-a VA + a -i^A-a
So, the desired result is
k2 -a)2 + l]
2=^[(2a2 +3)VA^ + .VA^],
where A = Vfl2 + 1.
SECTION 74
r cosxdx1. The problem here is to evaluate the integral — —= where a > b > 0. To do
_*.(*+
a
2X*+&)
this, we introduce the function f(z) =— —n~' whose singularities ai and bi lie
(z +a )(z +b )
inside the simple closed contour shown below, where R > a. The other singularities are, of
course, in the lower half plane.
130
According to the residue theorem,
eadx
jj? + a2)(x
2 +b2
where
and
That is,
or
B, = Res[/(zK<] = - -.v 2
*=<« (z + ai)(z +b2)
-+ J/(z)eft
£fe = 2»XA + ^)./ r.
52 =Res[/(z)e'
z
]=
Ise"dx n (e
b
_£1a
2 -b2\ b a
cos*dfr
(x2 + a
2)(x
2 + b2) a
2 -b2
( -b -a\
:=oi.2a(*
2 -a2)/
e-»
=bi2b(a
2 -b2)i-
- \Az)e*dz,
Rejf(z)ekdz.
Now, if z is a point on CR ,
l/(z)l<MR where M» =—= = = 5-fi
* (R2 -a2
)(R2 -b2
)
md\e iz
\=e~y <1. Hence
ReJc/(z)^&|<|j
Cj[
/(z)e'Vz
So it follows that
r cosjtd!*
<>MBnR =nR
(R2 -a2
)(R2 -b2
)
-» as R -» 00.
i (jc2 + a
2)(jc
2 + fc2) a
2 -&2l ft a
(a>b>0).
f COS fl.y
2. This problem is to evaluate the integral —= dx, where a > 0. The function
/(z) =——- has the singularities ±1; and so we may integrate around the simple closedz +1
contour shown below, where R > 1.
y
We start with
where
Hence
or
)-1-—dx+ jf(z)eiaidz = 2mB,
-R X +1C,
B = Res[f(z)eUK
] = ^-:*=" 1 J
z + i 2i
lOX
\—^dx = ne-a - \f{z)e
iaidz,
131
Since
we know that
and so
r cosax , _a _ r ,, . ,„T ,
J -rrr^ = ^ a -ReJ/(zy<fe,
\f{z)\<MR where L-
Re \f{z)e^dzicR
R2 -V
J
cosax
x2 + l
dx = ne~
That is,
r cosax , it—5 ox = —
<
J*2 +l 2(a>0).
4. To evaluate the integral \
XS^Xdx, we first introduce the function
I x2 +3
z2 +3 (z-zOiz-zJ
where z, = V3i. The point ztlies above the x axis, and zi lies below it. If we write
fi&P^M. where Hz) = ,
z - Zl Z-Zt
132
we see that zxis a simple pole of the function f(z)e
i2zand that the corresponding residue is
B1= (j>(zl )
= V3/exp(-2V3) _ exp(-2V3)
2V3i 2
Now consider the simple closed contour shown in the figure below, where R > VJ
.
y
Integrating f(z)e'2i
around the closed contour, we have
R ilx
jf—dx = 2mBl -if(z)e
i2idz.
Thus
j -^T^dx = Im(2^)-ImjcJ(z)e
i2idz.
Now, when z is a point on CR ,
l/(z)l<MR , where MR=— > as R -» °°;
and so, by limit (1), Sec. 74,
Consequendy, since
limf f(z)eiudz = 0.
R-*~JC,
JmjcJ(z)e
i2tdz\<\j
cnz)e
iUdz
we arrive at the result
|^<fa=^p(-2V3), or j™H.,fr = fexp(-2V3).JLJf +3 J„*+3 2
133
The integral to be evaluated isf*f^"* dx, where a > 0. We define the function
L x +4
/(z) = ~Tj7^' and» by computing the fourth roots of -4, we find that the singularities
^=V2ete/4 =l + i and z2 = V2V3 *'4 = V2V*'
V*'2 = (1 + i)i = -1 + i
both lie inside the simple closed contour shown below, where R > V2 . The other twosingularities lie below the real axis.
The residue theorem and the method of Theorem 2 in Sec. 69 for finding residues at simplepoles tell us that
R 3 tax- x e
where
and
f 7+4* +k f(z)eiKdz = 27n'^ +^
5 - Res^g* - zi
giaZ' - = g
'a(1+,)
= ggH1
t-tt z4 + 4 4z
34 4 ~ 4
o7 — Kes— = — = = =z=zlZ* + 4 4z* 4 4 4
Since
27n'(A+£2 ) = flu;. e +e
\
= i7te cos a,
we are now able to write
r x smax . r
J -^r^pfr = « "cosa-Im J /Uy*.
134
Furthermore, if z is a point on CR , then
R3
l/(z)l < MR where MR =——- -> as R -> <»;R — 4
and this means that
^LRfWehZdz
\ * \fCRmeiazdZ
according to limit (1), Sec. 74. Finally, then,
7 x 3sinax , .—3
—
—ax = ne cosaJ x +4
-» as /? -> °°,
(a>0).
8. In order to evaluate the integralf
*, we introduce here the function
J (x2 + l)(*
2+9)
f(z)= 2 2 ^ox - Its singularities in the upper half plane are i and 3i, and we
consider the simple closed contour shown below, where R > 3.
Since
Res[/(z)e*]=z
'f2(z+0(r+9)
1
16e
and
Res[/(z)eiz
]= 71
-«= 3'
1 J(z +
z3e*
(zz + l)(z + 3i)
Z=3iI6e
3 »
the residue theorem tells us that
t x3eadx r /'i
,( (^ + l)(^ + 9)+Jc/^ =H-I?16<? 16e
3 '
or
r j: sinxdx n ( 9 \ t
Now if z is a point on CR , then
135
\f{z)\<sMR where MR =—- —n
* * (R2-l)(R2 -9)
as R-> oo.
So, in view of limit (1), Sec. 74,
Im\cJ(z)e*dz\ <
|\cJ(z)e*dz ->0 as /?-> oo;
and this means that
"r _jc^nxdx_ _ n_(9__\°f
xi smxdx it (9
l(x2+\){x
2+9) 8e{e
2)' °T J (x
2 + l)(x2 +9)~ I6e \e
2
The Cauchy principal value of the integral f fnx<ix
can |,e foun(j ^ih^ f^J x
2 +4* + 5
function /(z) =1
r +4z + 5and the simple closed contour shown below, where R > V5.
Using the quadratic formula to solve the equation z2+4z + 5 = 0, we find that / has
singularities at the points zl=-2 + i and z
x=-2-i. Thus f(z) = 3—— , where z^
is interior to the closed contour and zxis below the real axis.
(z-zjiz-zj
The residue theorem tells us that
where
S = Res
and so
r smxdxI—5 r = Im
J„ x
l + Ax +
2me ,Zl
5 L(zi-^i).
-Imf f(z)ekdz,
136
or
r sinxdx % . „ , f
Now, if z is a point on Q, then \elz
\=e~y < 1 and
l/(z)l<M„ where MR=
Hence
Imf /(z)^|<|f
and we may conclude that
<MRnR = — 2-» as R -» oo,
(/?-V5)2
. 7 sin;t<£t # . „VJ~—
'a 7= sm2-
J x2 + 4x + 5 e
10. To find the Cauchy principal value of the improper integral f^x
.
+ 1^cosxdx, we shall use
J x2 +4x + 5
the function /(z) =z+l z+1
r + 4z + 5 (z-zJiz-zO— , where z
x= -2 + /, and z
x= -2 - 1, and the
same simple closed contour as in Exercise 9. In this case,
where
Thus
or
5 = Res(z + l)e*
(z-^Xz-zi)_ (Zt+l)g
fel
= (-l + Qg-2 ''
(z-zi) 2ez
r (x + Dcosjc . tt, . v r
Finally, we observe that if z is a point on Cs , then
R + l
\f(z)\<MR where Ms= R + l
(R-izMR-m (/?-V5)2-» as R -» oo.
Limit (1), Sec. 74, then tells us that
RcJCa
/(z)«*<fe|^|jCa/(z)e*&
137
-» as /? -> oo,
and so
_,, r (* + l)cos;t , n, . _pv - ^—: r^ =— (sin 2 -cos 2).
12. (aj Since the function /(z) = exp(i'z2) is entire, the Cauchy-Goursat theorem tells us that its
integral around the positively oriented boundary of the sector 0£r<R, 0<6<nl A
has value zero. The closed path is shown below.
A parametric representation of the horizontal line segment from the origin to the point
R is z = x (0<x<,R), and a representation for the segment from the origin to the point
Re'"* is z = re""4
(0 < r <, R). Thus
« A
jeixl
dx +jc
elll
dz-eiK'*\e-
rl
dr = <d,
or
)eixl
dx = ei
"'*]e-rl
dr-\c
ee
dz.
By equating real parts and then imaginary parts on each side of this last equation, wesee that
and
R 1*1jcos(x
2)dx = -j=je~
r*dr-Rej e*dzo
C*
R|
R
fsin(jc2)dbt = -H<f r2
</r-Imf e*dz.1 V2 i
Jc*
138
(b) A parametric representation for the arc CR is z = Reie(0 < < 7r / 4). Hence
it/* a/4
jce
izl
dz= \e*1'a'Rie
ied9 = iR\e-RlA* 26
eiRla"ie
ei6de.
= 1 and \eiB
\
= 1, it follows thatSince.iR
z cosZ0
, ic/4
Jce*dz<R je-*
lsia26dd.
Then, by making the substitution <j) = 2d in this last integral and referring to the form
(3), Sec. 74, of Jordan's inequality, we find that
2 2¥~4R-» as R -» oo.
(c) In view of the result in part (b) and the integration formula
oz
it follows from the last two equations in part (a) that
]cos(x2)dx =^^ and Jsin(*
2)d!* =
SECTION 77
1. The main problem here is to derive the integration formula
- cos(ax) - cos(foc) ,
2 dx = —(p - a) (a>0,b>0),
using the indented contour shown below.
y
Applying the Cauchy-Goursat theorem to the function
139
/(*) =
we have
or
f f(z)dz+[ f(z)dz+l f(z)dz+i f(z)dz = 0,JM JC* JL1 «C.
Since ^ and have parametric representations
Ll:z = re
i0 = r(p<r£R) and - L2 :z = re" = -r(p<r^R),
we can see that
^ wr ibr & —iar —ifer
^ /<«)& + jLif(z)dz = jj(z)dz-l_J{z)dz = \
S ~ edr +
\
e ~ edr
f (g^+g^ ) - (e*r + e"*') , „ J cos(ar) - cos(fcr) ,
J-5 dr =
2J-5 dr.
Thus
p
In order to find the limit of the first integral on the right here as p -» 0, we write
fiz) =\z
^ |
iaz|
(iaz)2
(
(/az)3
|
N
/(a-fc)
1! 2! 3!
+ ••• (0<lzl<°°).
(ibz(ibz)\(ibz)\
1H 1 H h •••
I 1! 2! 3!
From this we see that z = is a simple pole of f{z), with restdue~Ik = i(a - b). Thus
lim f f(z)dz = -B 7ti = -i(a - b)m = %{a - b).p—»0 JCp
140
As for the limit of the value of the second integral as R -» «>, we note that if z is a point onCR , then
rt ^ \e"*\+\eM
\e-v + e-* ^1 + 1 2
<—tkR =— -*0 as
R2 R
Consequently,
It is now clear that letting p -» and /?-><» yields
^7 cos(ar)-cos(frr) J
2Jj Ldr= n(b-a)
o
This is the desired integration formula, with the variable of integration r instead of x.
Observe that when a - and b = 2, that result becomes
7 1 - cos(2;t),
J3 dx = n-A
But cos(2jc) = 1 - 2 sin2x, and we arrive at
I * 2
2. Let us derive the integration formula
r x (l-a)TT, ,
|(7TIF'fa =
4cos(a,/2)
(-1<a<3)'
where x" = exp(aln x) when * > 0. We shall integrate the function
_ z" exp(alogz) . _ 3;^
whose branch cut is the origin and the negative imaginary axis, around the simple closedpath shown below.
p
Branch cut
By Cauchy's residue theorem,
f f(z)dz+i f{z)dz+[ f(z)dz+i f(z)dz = 2mResf(z).JL
iJL
zJC
p i=i
That is,
f f(z)dz+ f f(z)dz = 2mResf(z)~ f f(z)dz- f f(z)dz.
Since
Ll:z = re
i0 =r(p<r<R) and -I2 :z = refat = -r(p<r</?),
the left-hand side of this last equation can be written
* a(lnr+i"0) R o(lnr+/)r)
« a * a * a
= 171— dr + «"*\ ,i 2 <fr = (1 + f / . dr.
J (r2
-4- 1)2 J(r2
+1)2 'J(r2 + 1)
2
Also,
Res/(z) = where </>(z) =(z + 2 '
the point z = i being a pole of order 2 of the function f(z). Straightforward differentiation
reveals that
'(-\ _ -(a-l)logz0'(z) = e
a(z + i)-2z
L (z +3
142
and from this it follows that
Res f(z) = -ie-iaKllfl-a
We now have
Once we show that
limf f(z)dz = and liraf f(z)dz = 0,
p-*0 iC„ JC.
we arrive at the desired result:
„a _/i _\ JaKll -iaKll \_r ^ _ — a) _e e ^(1 - a)
Jo(^+l)2
(l-a)ff
2 l + em* eiaic —iaKll
4 eia*'2 + e
-ia*l2AcQS(a7t / 2)
The first of the above limits is shown by writing
I f(z)dz(l-p2
)
2r(1-p
2
)
np =Tip
a+l
and noting that the last term tends to as p -» since a + 1 > 0. As for the second limit,
1 _ 1
\jcmdz Ra
(tf2-l)
2 ""(R
2-l)
2 1R4 |1-
and the last term here tends to as /?-»<» since 3 - a > 0.
3. The problem here is to derive the integration formulas
"(Mxlnx i n2, . 7 VIf
-yxln* , jr . _ f Vjc , 7t
l x2 + l 62 J*2 + l V3
by integrating the function
f(r, z
mlogz _ e^logzH )
z2 +l ~
z2 + l
lzl>0,— <argz<—
around the contour shown in Exercise 2. As was the case in that exercise,
f f(z)dz + f f(z)dz = 2mResf(z) -f f(z)dz - f f(z)dz.
Since
/W-M where =z-i z+i
the point z = i is a simple pole of f(z), with residue
Res/(z) = 0(O = ye'*/6
.
The parametric representations
Ll:z = re
i0 =r(p<r<R) and -L1:z = reiK = -r(p<r<R)
can be used to write
J. and f =^.J^l^fr.
Thus
143
J^TT^ J r> + i
dr=Te-ic/^z-jc
mdz.f Vrlnr^ (
gl„/3
1Ifrlnr+in Vr ^ ;r
2
I r2 + 1 ;p p
By equating real parts on each side of this equation, we have
m^dr + cos(;r / 3)l^fdr - nsin{7t 1 3)\-^-dr =-—sin(;r / 6)
-Ref /(zVfe-Ref /(*)&;
and equating imaginary parts yields
* 3/~~ i * 3/~" 2
sin(;r / 3) f-^-f flfr + *rcos(w / 3)\——dr =— cos(;r / 6)
Jr+1 J r
2 +l 2
•Imf /(z)<fe-Imfr
/(z)<fe.
Now sin(^/3) =^, cos(^/3) = ^-, sin(7r/6) = ^-, cos(^/6) =— and it is routine toz Z 2 2
show that
limf f(z)dz = and limf /(z)<fc = 0.
p->0 Jc, ' /}_- Jc,v 7
144
Thus
That is,
3f V7lnr W3 r W . ;r2
2 * r2 +l 2 J r
2 +l 4
3 _ W3 _ n2
2l
22 ~ T*
21
22
4
Solving these simultaneous equations for /j and 72 , we arrive at the desired integration
formulas.
4. Let us use the function
v (logz)2
f, i n n I*/(z) =-2— lzl>0,--<argz<-
Z +1 V 2 2
and the contour in Exercise 2 to show that
f(lnx)
2
j n3
} Inx, n——-<& =— and -5— dtc = 0.
{x2 +l 8 JQx
2 + l
Integrating /(z) around the closed path shown in Exercise 2, we have
ff(z)dz + j
'
nz)dz = 2mResAz)-[ f(z)dz-\ f(z)dz.
Since
/W-M whereZ-i z+i
the point z = / is a simple pole of /(z) and the residue is
Res/(z) = 0(0 =^ =(lnl + t-;r/2)
2
= _*1*=•' 2/ 2/ 8i
'
Also, the parametric representations
Ll:z = re
i0 =r{p<r<R) and -L2 :z = reiic = -r(p<r<R)
enable us to write
J,/W*"f$£* and y^,]^fdr.
P PSince
p p pthen,
p p p
Equating real parts on each side of this equation, we have
p p
and equating imaginary parts yields
r ,
27c\^-dr = lmi f(z)dz-lm[ f(z)dz.
p
It is straightforward to show that
lim f f(z)dz = and lim f f(z)dz = 0.
Hence
and
Finally, inasmuch as (see Exercise 1, Sec. 72),
r dr _ %Jr2 + 1
=2'
we arrive at the desired integration formulas.
146
5. Here we evaluate the integral dx, where a>b>0. We consider theJ (x + a)(x + b)
function
.1/3 expl ^logz
(z + a)(z + b) (z + a)(z + fc)
and the simple closed contour shown below, which is similar to the one used in Sec. 77. Thenumbers p and R are small and large enough, respectively, so that the points z = -a and
z = -b are between the circles.
A parametric representation for the upper edge of the branch cut from p to R is z = rei0
(p < r < R), and so the value of the integral of/along that edge is
r exp:(ln/- + iO)
f_LL ^r = f £ ^o
(r + a)(r + b) {(r + a)(r + b)
A representation for the lower edge from p to is R is z = ren* (p<r< R). Hence the
value of the integral of/ along that edge from R to p is
*exp -(lnr + /2;r)
fL3 \ dr = -e
i2*n\
i(r + a)(r + b)dr.
According to the residue theorem, then,
where147
3=Res/(z)exp|jlog(-a)j exp|j(lna +
/;r)J gM3^-a+b a-b a-b
and
exp
B2 =Resf(z) = -
Z=-b
|log(-t) exp1(\nb + in)
i"nMb
-b + a -b + a a — b
Consequently,
• V7 ^-2«-^) .J/(^,
J/(^(\-e™)\ ^ dr = --
Now
L f(z)dz(a-p)(b-p) (a-p)(b-p)
and
/(z)&V* D 2*/?
21 n „< 2nR =— — •- >0 as
(R-a)(R-b) (R-a)(R-b) MR2
Hence
f , 2me i*/3e/a~-Mb) e™ 2^ -Mb)
o(r + fl)(r + ft)
r(1 - e''
2"3^ - *)'
<f'*/3
(e'*/3 - - *)
n(Ma-Mb) _ n{Ma~-Mb) = 2ar V^-Vfrsin(;r/3)(a-fc) ^/3~ 7~~
V3~" a-fc'
-y(«-*)
Replacing the variable of integration r here by x, we have the desired result:
(V* fc^n Ma'-Mb
{(x + a)(x + b) V3 a-b(a>b>0).
148
6. (a) Let us first use the branch
z-i/2 exp^--logz
z' + l z2 + l
lzl>0,--^<argz<^
and the indented path shown below to evaluate the improper integral
P
Branch cut
Cauchy's residue theorem tells us that
f f(z)dz+i f(z)dz + f f(z)dz+l f(z)dz = 2mResf(z),M -' (-* '^2 •'Cp Z=i
or
f /(z)<fe + J f(z)dz = 2mResf(z)- f /(z)rfz-f /(z)<fe.
Since
I^-.z^re' =r(p<r<R) and -l^.z-re^ = -r(p<r<R),
we may write
dr
1)
Thus
Now the point z = i is evidently a simple pole of f(z), with residue
exp[-|logij exp -I^inl + f.5
149
z~m
Res f{z) =—.J=< z + i 2i 2i 2i 2iVV2
Furthermore,
L f(z)dz. no it Jp n-TvTT =72^ as p->0Vp(i-p
2
) 1-p2
and
\\cm<k k4r n
Finally, then, we have
-> as /?->oo.
(1
VKr2 + l) V2'
which is the same as
r dbc _ it
J-vW+1) ~V2'
To evaluate the improper integral [-j=— , we now use the branch
JQ^x(x2 + l)
-1/2 expf-^-logz
zz + l
(kl>0,0<argz<2;r)
and the simple closed contour shown in the figure below, which is similar to Fig. 99 in
Sec. 77. We stipulate that p < 1 and R > 1, so that the singularities z = ±i are between
C„ and CR .
Since a parametric representation for the upper edge of the branch cut from p to R is
z = re' (p£r< R), the value of the integral of/along that edge is
«exp1
„_..r —(lnr + i'O) „
V^(r2 + l)
A representation for the lower edge from p to is R is z = re'1" (p<r<R), and so the
value of the integral of/ along that edge from R to p is
«exp ~(\nr + iliz)
- fL 2
2-U = -e-* dr =
f_i dr.
Hence, by the residue theorem,
R R 1
where
-1/2
B[=Res/(z) =
z + i
exp --log/ exp -i-flnl-H-*2l 2
-;«/4
2/ 2/ 2i
and
.-1/2
52= Res/(z) =^
*=-' z -
1
exp4iogH)1
expIf, , .3*— lnl + i
—
2V 2-i3*r/4
-2i -2i 2i
That is,
Since
and
J„Vr(r +1) J J
In p 2k Jp
VP(i-p2
) l-p2
2;cR 2;r-»0 as oo,
151
we now find that
J
i , e —e e +e edr — it = n
J V^(r2 + l)
= n = Kco&urvrWhen x, instead of r, is used as the variable of integration here, we have the desired
result:
7 dx _ it
SECTION 78
1. Write2r dd _ r 1 dz_ r dz
]5 + 4sin0~-k J z -z~l \ iz ~^2z2 + 5iz-2'
5 + 4
where C is the positively oriented unit circle \z\= 1. The quadratic formula tells us that the
singular points of the integrand on the far right here are z = -i/2 and z - -2i. The point
z = -il 2 is a simple pole interior to C; and the point z = -2i is exterior to C. Thus
? M artier'—w-u -wfi)-^.Jo5 + 4sin0 *=--42z
2 +5iz-2j L4z + 5i_L_ 1/2 U»7 3
2. To evaluate the definite integral in question, write
f dd _ r 1 dz _ t 4izdz
Jl + sin2 0~Jc ( 7 - 7
-i\2 '
iz ~Jc z*_6z2 + r1 +
2i )
where C is the positively oriented unit circle \z\= 1. This circle is shown below.
152
Solving the equation (z2
)
2 - 6(z2) + 1 = for z
2with the aid of the quadratic formula, we
find that the zeros of the polynomial z4 - 6z
2 + 1 are the numbers z such that z2 = 3 ± 2V2
.
Those zeros are, then, z = ±^3 + 2-^2 and z = ±V3-2>/2. The first two of these zeros areexterior to the circle, and the second two are inside of it. So the singularities of theintegrand in our contour integral are
z; =V3-2V2 and z1 =-zv
indicated in the figure. This means that
where
By = Res— 4iz _ 4izt
/
*-* z4-6z
2+ l 4^-12^ zf-3 (3 -2V2) -3 2-V2
and
B2= Res
4/z _ -4/z,
Since
the desired result is
*=-*.z4-6z
2+ l -4z
1
3 + 12z1 zf-3 2V2"
i \ 2k V22^(A-H52 ) = 2^|--^j = ±|.-l| = V2^
J l + sin2
7. Let C be the positively oriented unit circle lzl= 1. In view of the binomial formula (Sec. 3)
Jsin2" BdO =
I Jsin
2" ddd = ±J
1
r
( z-z2M 2i
In
dz
iz 22"+1
(-l)(-1)"/ Jc
-l\2«
dz
22"+ii^/c|(T) z2
"(-z
"1)iz"^
22"+1
(-l)"i ft ^ *
Now each of these last integrals has value zero except when k = n:
153
Consequently,
jcz
1
dz = 2m.
jsm2n ddd =1 (2w) !(-!)"2m = (2n)!
22»+i "
(n!)2 22fl
(«!)2
7T.
SECTION 80
5. We are given a function / that is analytic inside and on a positively oriented simple closed
contour C, and we assume that /has no zeros on C. Also, / has » zeros zt (& = l,2,...,n)
inside C, where each zk is of multiplicity mt . (See the figure below.)
The object here is to show that
To do this, we consider the kth zero and start with the fact that
f(z) = (z-zk )
m>g(z),
where g(z) is analytic and nonzero at zk . From this, it is straightforward to show that
zf'iz) _ mkz zg\z) _ mk(z-zk ) +mkzk zg\z) _ _ ,zg\z)
,mkzk— 1 — 1 -WlH 1 .
/(z) z-zk g(z) z-zt g(z) g(z) z-zk
zg'(z) zf'iz)Since the term — — here has a Taylor series representation at zk , it follows that
g(z)
has a simple pole at zk and that
Hz)
/(«)
An application of the residue theorem now yields the desired result.
154
6. (a) To detennine the number of zeros of the polynomial z6 - 5z
4 + z3 - 2z inside the circle
lzl= 1, we write
/(z) = -5z4
and g(z) = z6+z
3
-2z.
We then observe that when z is on the circle,
l/(z)l = 5 and ls(z)l<lzl6 + lzl
3+2lzl = 4.
Since \f(z)\>\g(z)\ on the circle and since /(z) has 4 zeros, counting multiplicities,inside it, the theorem in Sec. 80 tells is that the sum
Az) + g(z) = z6-5z*+z3
-2z
also has four zeros, counting multiplicities, inside the circle.
(b) Let us write the polynomial 2z4 - 2z
3 + 2z2 - 2z + 9 as the sum /(z) + g(z), where
/(z) = 9 and g(z) = 2z4-2z
3 + 2z2-2z.
Observe that when z is on the circle lzl= 1,
l/(z)l = 9 and l#(z)l<2lzl4+2lzl
3 + 2ld2+2ld = 8.
Since l/(z)l>l#(z)l on the circle and since /(z) has no zeros inside it, the sum
/(z) + g(z) = 2z4 - 2z
3 + 2z2 - 2z + 9 has no zeros there either.
7. Let C denote the circle I zl= 2
.
(a) The polynomial z4+ 3z
3 + 6 can be written as the sum of the polynomials
/(z) = 3z3
and $(z) = z4+6.
On C,
l/(z)l = 3lzl3= 24 and l^(z)l = lz
4+6l^izl
4 + 6 = 22.
Since l/(z)l>l#(z)l on C and /(z) has 3 zeros, counting multiplicities, inside C, it
follows that the original polynomial has 3 zeros, counting multiplicities, inside C.
(b) The polynomial z4 - 2z
3 + 9z2 + z - 1 can be written as the sum of the polynomials
/(z) = 9z2
and g(z) = z4-2z
3+z-l.
On C,
|/(z)| = 9lzl2= 36 and U(z)l=lz
4-2z
3 + z-ll<lzl4+2lzl
3+lzl+l = 35.
155
Since l/(z)l >\g(z)\ on C and /(z) has 2 zeros, counting multiplicities, inside C, it
follows that the original polynomial has 2 zeros, counting multiplicities, inside C.
(c) The polynomial z5 + 3z
3 + z2 + 1 can be written as the sum of the polynomials
/(z) = z5
and $(z) = 3z3+z
2 + l.
On C,
l/(z)l=lzl5 = 32 and I#(z)l = l3z
3+z
2+ll<3lzl
3+lzl
2+l = 29.
Since \f(z)\>\g(z)\ on C and /(z) has 5 zeros, counting multiplicities, inside C, it
follows that the original polynomial has 5 zeros, counting multiplicities, inside C.
10. The problem here is to give an alternative proof of the fact that any polynomial
P(z) = a + a,z + • • • + a^z""1 + aHz" (a. * 0),
where n > 1, has precisely n zeros, counting multiplicities. Without loss of generality, we
may take an=l since
P(z) = a„
rEa. + BLz + ... + EB=L^ +z^
Let
/(z) = zB
and g(z) = a +o1z + - + an_l
z"'1
.
Then let R be so large that
/?>l + la l + lo1
l + ---+laB_ 1l.
If z is a point on the circle C: lzl= R, we find that
lg(z)l fS la l + l^llzl + ••• + la^llzl"-1 = la l + iaJR + ••• + la^ltf-
1
< Iflol/?"-1 + Ifl!!/?"
-1+ ••• + t^.J/?"
-1= (la l + laj + - +
</?/r-1 =/2"=lzr = l/(z)l.
Since /(z) has precisely n zeros, counting multiplicities, inside C and since R can be made
arbitrarily large, the desired result follows.
156
SECTION 82
1. The singularities of the function
2s3
5-4
are the fourth roots of 4. They are readily found to be
or
V2, <j2i, -V2, and -V2*.
See the figure below, where y > -4l and R > V2 + y.
Y+iR
The function
has simple poles at the points
2?V'e =——
-
s -4
and
j = V2, jt= V2/, j, = -V2, and *3
=
V 3 o„3*.r
X Res^FW] =2 Res ^il =f 2j£
2 2 2 2
• +
= coshV2f + cos V2f.
Suppose now that s is a point on CR , and observe that
157
\s\=\y+Re'°\ <y+R = R+ y and \s\=\y+Re ie\>\y-R\ = R-y>«j2.
It follows that
\2s3
\ = 2\s\3
<2(R+yf
and
Consequendy,
This ensures that
I s4 - 41 > lid
4 -4I> (/? - y)4 - 4 > 0.
lF(*)l< ^(J?+
.
y)'
->0as/?->(/?-y)
4 -4
/(/)= coshV2r + cosV2f.
2. The polynomials in the denominator of
F(s)2s -2
(s + l)(s2 + 2s + 5)
have zeros at s = -1 and s = -1 ± 2i. Let us, then, write
e"F(s) =e*(2j-2)
(j + IXj-^X^-J!)
where jt= -1 + 2/'. The points -1, su and s
{are evidently simple poles of e"F(s) with the
following residues:
i_ gJt(2^-2)
-Si)^=Resfe"F(j)]
z=-l L J
52= Res[*"F(,)l =
g"<(2^- 2) = fi - 1W<,
B3=Res[e«F(s)]=
^'^^3
~, L WJ(J, +1X51-5.)
eSl'(2Sl -2)
(5,+lX^-J!)
158
It is easy to see that
i2t
-t , -t— —e + e
f ei2'-e-
i2'
+en
' + e-i2''
= <f'(sin2f+ cos2f-l).v 2/ 2
Now let s be any point on the semicircle shown below, where y > andR > -\/5 + y.
Since
\s\=\y + Re ie\<y+R = R+y and \s\=\y+ Rei6\>\y- R\= R- y> V5,
we find that
I2j-2I<2IjI+2<2(/?+7) + 2,
and
\s + l\>\\s\-\\>(R-y)-l>0,
\s2 +2s + 51=1 s - Sl \\s - s,\ > (\s\-\sj)
2 > \{R - 7)2 - V5 f > 0.
Thus
Im = ^=1 < . Y) + 2
Is + II \s2 + 2j + 51 [(* - y) - !][(/?
- 7)
2 - V5]
and we may conclude that
/(/) = e"'(sin It + cos It - 1).
as R -> oo,
4. The function
159
„2 2s —a(r+a2
)2 (a>0)
has singularities at s = ±ai. So we consider the simple closed contour shown below, where
Y>0 and R>a+y.
Upon writing
F(s) = yv. 2
where 0(s) = -
(s - a/)
we see that is analytic and nonzero at s = ai. Hence sQ is a pole of order m = 2 of
F(j). Furthermore, F(s) = at points where F(s) is analytic. Consequently, s is also
a pole of order 2 of F(j); and we know from expression (2), Sec. 82, that
Res [e"F(s)] + Res [e^Fis)] = 2Re[eia,
(fc1+ b
2t)] ,
where and b2 are the coefficients in the principal part
s - ai (s - ai)2
of F(s) at ai. These coefficients are readily found with the aid of the first two terms in the
Taylor series for </>(s) about s = ai:
1
(s - ai)(s) =
1
(s - ai)
160
</>(ai),0'(ai)=^)r + 7^- + - (0 <!,-«!< 2a).
It is straightforward to show that <f>(ai) = 1 / 2 and <j>\ai) = 0, and we find that ^ = and
b2=l/2. Hence
Res [e*F(sj\+ Res [e"F(s)] = 2Re ^(j'j = tcosat.
We can, then, conclude that
f{t) = tcosat(fl > 0),
provided that F(s) satisfies the desired boundedness condition. As for that condition, when
z is a point on CR ,
\z\=\y+Reie\£y+R = R+y and \z\=\y+Re
ie\>\y-R\= R- y > a;
and this means that
\z2 -a2
\<\z\2 +a2 <(R+y)2 + a
2and \z
2 +a2\>\\z\
2 -a2\>(R-y)2 -a2>0.
Hence
6. We are given
„. . sinh(xs)
which has isolated singularities at the points
-n „ _ (2n-l)^ . _ (2n-l)n., , „= °> *.= 2 ,f 311(1 J
"= ~
2*
(n = l,2,...).
This function has the property F(s) = F(s), and so
f(t) = Res [e*F(sj\ + J^Res [«*F(j)] + Res [e"F(j)]j.
To find the residue at j = 0, we write
sinh(xy) _ xy + (xy)3/ 3 ! + • • • oc + jcV/6 + —
j2cosh* 5
2(l + *
2/2! + ...)" s + s
3/2 + - { 2
0<ld<-|.
161
Division of series then reveals that s is a simple pole of F(s), with residue x; and,
according to expression (3), Sec. 82,
Res le^Fis)] = Res F(s) = x.s=s„ L 1 s=s
As for the residues of F(s) at the singular points a; (n = 1,2, ...), we write
F(s) = - where = sinh(jcs) and = s2cosh j .
We note that
. . . . (2n-l);a _ , v „= isin^—^— 5*0 and q(sn ) = 0;
furthermore, since
q'(s) = 2j cosh j + j2sinh a,
we find that
„ . (2n-l)V. . (2w-1)tf .(2n-l)27r
2. ( n\?W = ism = -,i -J. sm\nn--
J
.(2n-l)2^V. jt , art (2n-l)
2 ^2,
' n-r smnttcos— -costt7rsin— = v — (-l)"/*0.4 V 2 2) 4
In view of Theorem 2 in Sec. 69, then, sn is a simple pole of F(s), and
?'(*.) (2n-l)J
2
Expression (4), Sec. 82, now gives us
Res He. [e"F(s)] - 2 Re(-± .-^sini^^expTifi^l]'='.
1 J'='"»
L J[;r
2(2n-l)
22 L 2 JJ
8 (-1)" . (2n-l)*ct (2n-l)nt= —?•——^—rsin- -—cos- —
.
n1(2«-l)
22 2
162
Summing all of the above residues, we arrive at the final result:
f(t) - x+— 2_,— -j sm cos- —
,
7. The function
1F(s) = -
JCOSh(51/2
)
'
where it is agreed that the branch cut of sm
does not lie along the negative real axis, has
isolated singularities at s = and when cosh(jl/2
) = 0, or at the points s„ = _(2n Jj n
in = 1,2, ...). The point s is a simple pole of F(s), as is seen by writing
.ycosh(.y1/2
) 4l + (j1/2
)
2/2! + (/
/2
)
4/4!+-] s + s
2 /2 + s3 I2A+-
and dividing this last denominator into 1. In fact, the residue is found to be 1; and
expression (3), Sec. 82, tells us that
Resfe"F(s)] = Res F(s) = l.
As for the other singularities, we write
F(s) =^ where p(s) = 1 and q(s) = scosh(sm
).
Now
p(sn ) = 1*0 and q(sn ) = 0;
also, since
q\s) = ±sm sinh(51/2
) + cosh(*1/2
),
it is straightforward to show that
x (2n-l);r . ( n\ (2n-V)7t4 (*„) = -^-^-sinl nit -- I = ±J±-^L(-Vf±^
So each point sn is a simple pole of F(s), and
ResF(J)= pw = i.i=«l# CO K 2n-l
163
Consequently, according to expression (3), Sec. 82,
Res \e"F(s)] = e'"' Res F(s) =- • £-^-exp'=', L J
;r 2n -
1
Finally, then,
or
(2n-l)Vf
/(/) = Res [estF(s)] +£ Res [**F(j)1 ,
nZi 2n-l
(2n-l)Vf
(n = l,2,...).
Here we are given the function
. . _ coth(^y / 2) _ cosh(^y/2)
s2 + l ~(/ + l)sinh(^s/2)'
which has the property F(s) = F(J). We consider first the singularities at s = ±i. Upon
writing
\ u jl /• \ coshers/ 2)F(j) = where = -—
,
s-i (s + i)sinh(ns f 2)
we find that, since 0(i) = 0, the point i is a removable singularity of F(s) [see Exercise
3(b), Sec. 65]; and the same is true of the point -i. At each of these points, it follows that
the residue of e"F(s) is 0. The other singularities occur when nsl2 = nni
(/i = 0,±l,±2,...), or at the points s = 2ni (n = 0,±1,±2,...). To find the residues, we write
F(s) = £yl where p(s) = coshf—Jand q(s) = (s
2 + l)sinh[—q(s) \2 J \2
and note that
p(2ni) = cosh(nm) = cos(nn) = (-1)" * and q{2ni) = 0.
Furthermore, since
q\s) = (s2 +
1)|cosh^H
j+ 2,sinh(-0
we have
q\2ni) = {-An2 + l)-^cosh(n^i) = (-4n 2 + l)-cos(n^) =^ 2
/r(4w2 - 1)
(-1)"*0.
Thus
*=2 »<- q\2ni) n An2 -I(« = 0,±1,±2,...).
Expressions (3) and (4) in Sec. 82 now tell us that
Res[e"F(5)l = ResF(1y) =-
and
Res[e"F(j)]+ Res \e"F(sj\ = 2Re
The desired function of t is, then,
ei2"'(
n An 2 -I
A_ cos 2nt
n An2 -\
*r*\ 2 4^cos2/if
(« = 1,2,...).
The function
F(V) -sinhjxs
1'2
)K
' J2sinh(^
1/2
)
(0<*<1),
where it is agreed that the branch cut of sm
does not lie along the negative real axis, has
isolated singularities at « = and when sinh(jl/2
) = 0, or at the points s = -n 2it
2
in = 1,2, .. .)• The point s = is a pole of order 2 of F(s), as is seen by first writing
sinh(x?1/2
) _ xsm+ (xs
U2)
31 3! + (xs
m)5/ 5! +• • • x + x 3
sl 6 +xV 1 120 +• •
s2smh(s
m) s
2[s
1'2 + {smf / 3! + (s
m)51 5! +•••]
~s2 + s
31 6TP 1 120 +•••
and dividing the series in the denominator into the series in the numerator. The result is
- =x— + -(x -x]s
sinh(xy1/2
) _ 1 13 1
2 • . / 1/2 \— X~T + ~T\X ~X)—h'
rsinhCs1 '2
) s2
6
165
(0<\s\< 7T2).
In view of expression (1), Sec. 82, then,
Res [e"F(s)] = Ux'-x) + xt = \x{x% - 1) + xt."°° 1 1 6 6
As for the singularities s - -n2n2(n = 1,2, ...), we write
F(s) =4t where = sinh(^1/2) and q(s) = s
2sinh(j
1/2).
q(s)
Observe that p(-n2n2)*0 and q{-n
2n2) = Q. Also, since
q'is) = 2s sinh(51/2
)+ ijj 1/2cosh(j
1/2
),
it is easy to see that q'(-n2n2
) * 0. So the points s = -n2^2(n = 1,2, ...), are simple poles
of and
>=-« 2jtj
£ (j)
2sinh(xs1/2
)
™1/2cosh(5
1/2
)
2 (-1)"+1
= —5- 5—sinn;cc (/i = 1,2,...).
.1-1 * «
Thus, in view of expression (3), Sec. 82,
2 (-1)B+1
Finally, since
Res le"F(s)] =4 • ^f-*""V'^nnx
f(t) = Rcj[e^F(s)] +±ReAyF(S)],
we arrive at the expression
fit) = ~x(x2-l) + xt —1
—
g " ***
sin6 jr - 3
2v(-ir',,v,,n
(n = l,2,...).
10. The function
F00=——
-
5 jsinhj
has isolated singularities at the points
s = and j„ = nm, s„ = -nm (n = 1,2,. . .).
166
jsinhj = 5| s + —s 3+---) = s
2 + —s4+-
Now
jsinhj = 5i s +6 ) 6
and division of this series into 1 reveals that
in 1
(0 <ld< oo),
(0<\s\<7t).
This shows that F(s) has a removable singularity at s . Evidently, then, es'F(s) must also
have a removable singularity there; and so
Res \es'F(s)] = 0.
s=s L J
To find the residue of F(s) at sn = nm (n = 1,2,. . .), we write
^0) =-— where p(.y) = sinhj-j and = ,y
2sinh.s
and observe that
p{nni) = -nm * 0, q(nm) = 0, and q'(nm) = n27t
2(-l)
n+l*0.
Consequently, F(.s) has a simple pole at sB , and
Res F(.) =4^ =-rrar =^ 0» = U,...).tf'(n;n) nV(-l)n+1
n;r
Since F(j) = F(s ), the points sn are also simple poles of F(s); and we may write
Res \estF(s)] + Res \e"F(sj\ = 2Re ( lT
ieinn = 2Re
Ml
(-1)"(icos/i7#-sin/Mtt)
= 2 sinnnt.nit
Hence the desired result is
/(0 = Res [e"F(sj\ +f (Res [e*F(j)l + Res [^'F(.y)l},
or
-smnm.
167
11. We consider here the function
„. . sinh(xj)F(s) =—
2 h . (0<x<l),s(s +co)coshs
where a > and cd*G)„ =— (n = 1,2,...). The singularities of F(j) are at
5 = 0, 5 = ±tui, and ^ = ±6)J (n = l,2,...).
Because the first term in the Maclaurin series for sinh(xs) is xs, it is easy to see that s = is
a removable singularity of e"F(s) and that
Res[«*F(j)l = 0.s=s 1 1
To find the residue of F(s) at s = coi, we write
cv x u jl v sinh(xy)F(5) =^^: where <t>(s) =- ;
',
s — coi s{s + cm)coshs
from which it follows that s = coi is simple pole and
t> rv \ ^/ -\ sinh(jcflw) /sin oxRes F(s) = </>(0)i) =——
7
coilcoi cosh(o)i) -2co coso
Since fjs) = F(J), then,
Res \e"F(s)]+ Res [e*F(5)l = 2Re|" jSi"^ iVl =
lj
lj L-2tf) cos© J
- sinfitt sin oar sin g»2—= sin G)f =
2fiJ2COSfl) G)
2cosg>
As for the residues at s = ©„/ (n = 1,2,. . .), we put F(s) in the form
F(s) = 2^- where p(j) = sinh(*.y) and q(s) = (s3 + G)
2s)coshs.
q(s)
Now p(fl)„i) = swh(x(Q„i) = i sin fi)„.x * and q(0)„i) = 0. Also, since
q'(s) = (s3 + 0)
2s)sinhs + (3s
2 + fiJ2)cosh s,
we find that
q'(o)J) = (-coli + Q)2G)J)smh(o)ni) = -Q)„ (m
2 - <a2)sin con * 0.
Hence we have a simple pole at s = ani, with residue
q'(o)ni) -con(a)
2 - G)2)sm G)n
168
Consequently,
Resfe*F(s)]+ Res \e"F(s)] = 2Re-G)n ((Q
2-G)
2
n )sin(Dn
= 2-smw.xsinco.t
0)n (a}2 -Q)2
„)smo)n
But sin o)„ = sin( n;r - -j j= (-l)"
+l, and this means that
Res[e"F(*)l + Resfg'T(j)1 =2^^>^*sin6)"'
j-*.! 1- > 6>_ a)
2-co
2
Finally,
/(f) = Res [e"F(s)] + (Res[e"F(*)] + Res [e"F(*)]J + J)fRes [e"F(*)] + Res [e"F(s)]\
That is,
_ sinffit*sin6tt[ 2y sina)„;csinfly
CO COSO) ^ 0)n © - a:
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