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8/10/2019 Comp Ana SRJ
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1. Dr. Prof. S.R. Jayaram
Department of Mathematics
SBM Jain College of Engineering,Jakkasandra 562112, Bangalore rural dt
(Please give your feed back to [email protected])
2.Complex analysis
Complex number
a +ib, i=-1
Exponential form
x+iy = r{cos-1
(y/x )+i sin-1
(y/x)}, r = (x2
+y2
)
Eulers formula
eix = cosx + isinx
Representation of complex number
-Argonds plane, Complex plane
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5.Complex numbers
Adding (a+ib) (c+id) = (a+c) i(c+d)
Multiplication
(a+ ib) x (c + id) = (ac - bd) i(bc+ad)
Division
(a + ib)/(c + id)
= [(a + ib)/(c + id)]x (c-id/c-id)
= [ac+bd) +i(bc-ad)]/[c2
+ d2
]
6. Complex numbers are not ordered
Equality (a+ib) = (c+id) iff a =c and b =d
i= -1, i2 = -1, i3 = -i, i4=1
1/i =-i, 1/i2 = -1,1/ i3 =-i, 1/i4=1
In exponential form only principal value or periodic function is taken
7 Functions- complex
Single valued
f(z) = f(x + iy), i = -1
Differentiability
CR equations ux
=vy
, uy
= -vx
f(z) = ux + ivx or -iuy - vx
f(z) is analytic
f(z) = Logz, ez, z+1/z, z
2,
1/2log(x2+y
2) +itan
-1y/x
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11. Complex analysis
If u(x,y) = 2xy + xy2 -2y3, is u harmonic?
Ux =2y + y2
, Uxx
= 0
Uy =2x + 2xy -6y2
Uyy
= 2x12y
Uxx+ Uyy 0, so U is not harmonic
12.Complex Analysis
Show V = -sinx sinhy is harmonic. Find its conjugate U
Answer: Vx = -cosx sinhy, Vxx = sinx sinhy, Vy = -sinx coshy, Vyy = -sinx sinhy
So Vxx + Vyy = 0
From CR equations, Ux = Vy and Uy = -Vx
So u(x,y) = cosx coshy + constant = -Vx
So u(x,y) = cosx coshy
f(z) = cosx coshy +c +i(-sinx sinhy)
13. Complex Analysis
Find analytic function, given
u = ex
(xcosy-ysiny) + 2sinxsinhy+ x3
- 3xy2
+ y
Ux = ex(xcosy-ysiny) + excosy +
2cosx sinhy + 3x23y2
Uy =ex
(-xsiny-sinyy cosy) + 2sinx coshy
-6xy +1
f(z) = Ux + i Vx = Ux - i Uy = zez+e
z+3z
2-2isinz-i replacing x by z and y by zero)
Integrating, f(z)= zez+ez+z3/3 -2icosz-iz +c
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14. Complex analysis
If V = (x2y2 )+ x/(x2 + y2 ), find u
V is harmonic, Vxx
+ Vyy
= 0
Vx = -2y2xy/(x2
+ y2
)2,
= Uy
Integrating U = -2xy + y/ (x2 + y2 ) +c(y), where c(y) is a function of y
Now u = -2xy + y/ (x2 + y2 ) +c
15 Complex analysis
Given u = sin2x/(cosh2ycos2x) find v
Let f(z) = u + iv
f(z) = ux
+ i vx
= vx
- i uy
= (2 cos 2x cosh 2y2 )/ (cosh2ycos2x)2
- i (2 sin 2x sinh 2y )/ (cosh2ycos2x)2
=By Milne Thompson method,
f(z)= (2cos2z-2)/ (1-cos2z)2 + i(0)
= -cosec2z. So f(z) = cot z + c
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16.Complex analysis
Find the analytic function given
v = r2
cos 2 r cos + 2
r ur = v = -2 r 2 sin 2 + r sin
-1/r u = vr = 2r cos - cos , so
ur
= -2r sin 2 + sin , Integrating w.r.t r
u = -r2
sin 2 + rsin + c
So f(z) = i(r2
e2ir ei) + c + 2i
17 Complex analysis
If f(z) is regular, prove
(uxx + uyy)|f(z)|2 = 4 |f(z)|2
f(z) = u + iv, so that |f(z)|2 = u2 + v2 = (x,y)
x = 2uux + 2vvx and
xx = 2[ uuxx + (ux)2
+ vvxx +(vx)2
], similarly
yy = 2[ uuyy + (uy)2
+ vvyy +(vy)2
],
So xx+yy= 2[u(uxx+uyy ) + v(vxx + vyy)]+ 2[(ux )2
+(u y)2]+ (vx )
2+(vy)
2]
By CR equations, (ux )
2
= (vy )
2
, (u y)
2
= (-vx )
2
And uxx+uyy = 0, vxx+vyy = 0,
so xx+yy= 4[(ux )2
+(u y)2 ] = 4|f(z)|2
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18. Complex analysis
Find if uv = (cosx + sinxe-y)/2(cosxcoshy), f(/2)= 0
Ux- Vx= [(sinx-cosx)coshy + 1e-y sinx]/
2[cosxcoshy)2 (1)
Uy- Vy= [(cosx - coshy)e-y
+
(cosx + sinxe-y )sinhy]/2[cosxcoshy)2
-Vx- Ux= [(sinx + cosx)sinhy + e
-y(cosxcoshy -sinhy]/2[cosxcoshy)2
(2)
Subtracting (2) from (1) and adding (2) and (1)
And putting x = z, y = 0, we get
19. Complex Analysis
f(z) = Ux
+ iVx
= (1- cos z) /2(1- cos z)2
Integrating, f(z) = -1/2 cot (z/2) + c Using initial condition that f(/2) = 0
We get c =
So f(z) = ( 1cot z/2)
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20 Complex Analysis
Given v = (r -1/r)sin, find u
CR equations are u
r
= 1/r v
and u
= -r v
r u = -r vr = -(r + 1/r) sin , Integrating
u = (r + 1/r) cos + c(r), Differentiating
Using CR c(r) = 0, so u = (r + 1/r) cos
f(z) = (r + 1/r) cos + i(r - 1/r) sin + c
21 Transformations
1. Translation w = z + c
Taking z = (x + iy) and c = (a + ib)
w = c + z = (x + a) + i( y + b), so a point
(x, y) is mapped onto (x + a) + i( y + b).
The object is moved through a distance c.
Image retains same shape and size
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24 CA-Inversion and reflection
W = 1/z
Let z = re
iand w = Re
i
Then R = 1/r and = -
25 . CA-Bilinear transformation
W = (az + b)/(cz + d), a, b, c, d are complex constants and ad - bc 0. The
condition ad - bc 0 implies dw/dz 0.
The inverse transformation is given by
z = (-dw + b) / (cw - a)Fixed points of transformation is the solution
of cz2
+ (d-a) zb 0
26. CA-bilinear transformation
By actual division w = (az + b)/( cz + d) is
w = a/c + (bcad)/c2
. 1/(z+d/c)So consider w1 = z+d/c, w2 _= 1/w1
And w3
= (bc - ad)/c2
w2
w = a/c + w3
Translation, rotation, magnification
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27 CA- Bilinear transformation
The cross ratio of four points z1 , z2 , z3 , z4
is denoted by (z1 , z2 , z3 , z4 ) and is defined by (z1 - z2 )(z3 - z4)/ (z1 - z4 )(z3 - z2)
Bilinear transformation preserves cross-ratio of any four points
w1 = (az1 + b)/(cz1 +d), w2 = (az2 + b)/(cz2 d),
w3
= (az3
+ b)/(cz3
+d), w4
= (az4
+ b)/(cz4
+d),
28 CA-BLT
(z1
- z2
)(z3
- z4)/ (z
1- z
4)(z
3- z
2) =
(w1
- w2
)(w3
- w4
)/ (w1
-w4
)(w3
- w2)
So the bilinear transformation preserves cross ratios of any four points
29 Complex Analysis
Let us denote four complex constants a, b, c, d and define the function
f(z) = w = (az + b)/(cz+d) is called Mobius/bilinear transformation
There exists a unique transformation which maps three points z1,
z2,
z3 to the
points
w1
,w
2,w
3given by the formula
(z- z1)(z2-z3)/(z-z3)(z2-z1) =
( w - w1)(w2 - w3)/(w - w3)(w2 - w1)
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30 CA-Bilinear transformation
(-1, 1, i) to (-1, 0, 1)
W = i(1-z)/(1+z)
(-2, -1-i, 0 ) to (-1, 0, 1)
W =[(1- i)z+2]/[(1+i)z+2]
(1, i, -1) to (i, 0, -i)
W = (1+iz)/(1-iz)
(1, i, -1) to (i, 0, -i)
W = z(6+3i)+(8+4i)/ z(7+6i)+(6-2i)
31 CA-Bilinear transformation
Find the invariant points of the BLT
w = (z-1)/(z+1), z =i
W = (6z-9)/z, z = 3 Find BLT which maps (-1, 0,1) to (0, i, 3i)
w = -3i(z+1)/(z-3),
Find BLT which maps (1, 0,-1) to (i, 1, )
w = (-1+2i)z+1)/(z+1)
Find BLT which maps (0, 1, ) to (-1,- i, 1)
w = (z-i)/(z+i)
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32 Complex Analysis-Integration
Line integrals
c (u+iv)(dx+idy) =
c (udx-vdy) + ic (vdx+udy)
Properties of line integrals
Linearity
ac f(z)dz + bc g(z)dz =c (af(z)+ bg(z))dz
Sign reversalc f(z)dz = -c f(z)dz, where the c traversed in opposite direction
Partitioning of path c f(z)dz =c1 f(z)dz +c2 f(z)dz
33 Line integrals
Evaluatec |z|2dz around the square with vertices (0,0), (1,0), (1,1), (0,1)
The square is enclosed by four lines c1, c2, c3, c4
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34 Line integrals
cf(z)dz=c1f(z)dz +c2f(z)dz +cf(z)dz +c f(z)dz
Along c1, y=0,c1 |z|2dz =c1 (x2+y2)(dx+idy) =01 x2 dx = x3/3 = 1/3
Along c2, x = 1, y varies from 0 to 1
01
(1+y2
)idy = i4/3
Along c3, y =1, and x varies from 1 to 0
10
(x2
+1)dx = -4/3
Along c4, x=0, y varies from 1 to 0
10 y2 idy = iy3/3 = -i/3
cf(z)dz = -1+i
35 Line integrals
c dz/(z-a) = 2i where c is the circle |z-a| = r
The parametric equation for C is z-a= rei
where varies from 0 to 2
dz = irei d
c
dz/(z-a) =c
(1/ rei
)irei
d = 2i
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36 Line integrals
Evaluate (x2
+iy)dz, along the paths y =x and y =x
2
for z varying from 0 to 1+ i
dz = dx + idy, so (x2 +iy)(dx + idy) =
Along y = x, (y2 +iy)(dy +idy) = (1+i)(y3/3+iy2/2)
For y varying from 0 to 1
= (1+i) (1/3 + i/2)
= (1/3 -1/2) + i(1/3 + ) = -1/6+i/6
37 Line integrals
If f(z) is analytic function and f(z) is continuous at each point within and on a
closed curve C then
c f(z)dz = 0
Writingc f(z)dz =c (udxvdy) + ic (vdx + udy)
Since f(z) is continuous ux
uy
vx ,
vy
are continuous in the region D enclosed by
C, Using Greens theorem,
c f(z)dz = -D [vx + uy] dxdy + i-D[ux -vy]dxdy. Since f(z) is analytic,
vx = - uy ,ux = vy and the two double integrals vanish identically.
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38 Line integrals
If f(z) is analytic function and if a is any point within C,
then f(a) = 1/2ic f(z)/(z-a)dz
Proof: Function f(z)/(z-a) is analytic, at all points of C except at z = a, We draw a small
circle C1 of radius r lying entirely within C. Now f(z)/(z-a) is analytic in the region
enclosed by C and C1 , by Cauchys theoremc f(z)/(z-a)dz =c1f(z)/(z-a)dz
For any point on C1 , z-a = rei
and dz = irei
d
c1f(z)/(z-a)dz =c1f(a+ rei )/ rei .i rei d
= ic1f(a+ rei )d . In the limiting case as r0,
ic1f(a)d= 2i, thus f(a) = 1/2ic f(z)/(z-a)dz
39 Similarly f(a) = 1/2ic f(z)/(z-a)2dz ,
f(a) = 2!/2ic f(z)/(z-a)3dz ,
f(a) = 3!/2ic f(z)/(z-a)4dz
.
fn(a) = n!/2ic f(z)/(z-a)n+1dz
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40 Cauchys integral formula
Evaluatec(z2
z+1)/(z-1) dz where C is the circle1. |z| = 1.
Answer: Given f(z) = (z2z+1) and a =1.
Since f(z) is analytic within and on |z| = 1and
a = 1 lies on C. By Cauchys integral formula
1/2ic f(z)/(z-1)dz = f(a) =1
2. |z| = . Lies inside the circle |z| = 1. Hence by
Cauchys theoremc(z2z+1)/(z-1) dz =0.
41 Cauchys integral formula
c(e2z
/(z-1)(z-2) dz, where C is the circle |z| = 3
Answer: f(z) = e2z
is analytic within the circle |z|=3,
a = 1 and a = 2 lie inside |z| = 3. Hence
c(e
2z/(z-1)(z-2) dz =
c(e
2z[1/(z-2) -1/ (z-1)] dz
=c(e
2z .1/(z-2)dz -
c(e
2z1/(z-1)dz
= 2ie4 - 2ie2 = 2ie2 (e2 -1)
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42 Cauchys integral formula
c(cosz/(z2
-1)dz, around a rectangle 2 i, -2 i
Answer: f(z) = cos z is analytic in the rectangle and the points z = 1, and z = -1, lieinside the rectangle.
c(cosz/(z2 -1)dz =
c[(cosz/(z -1)dz -(cosz/(z+1)dz ]
= [ 2icos(1) - 2icos(-1)]
= 0.
43 Cauchys integral formula
Taylors series: If f(z) is analytic in a circle C with center a, then for z inside C
f(z) = f(a) + f(a) (z-a)+ f(a) (z-a)2 /2! +..
+ +fn (a) (z-a)n /n! +.
Laurents Series: If f(z) is analytic in a ring
shaped region R bound by two concentric circles
C1 and C2 of radii r and r1 with r > r1 center a,
then for all z in R
f(z) = a0 + a1(z-a) + a2(z-a)2 +..+ a-1(z-a)
-1+
+ a-2(z-a)-2 +..+ where an = 1/2if(t)/(t-a)n+1 dt
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44 Taylors, Laurent Series
1. Find Taylors series expansion of 1/(z+1)2 around z = -i
Answer: f(z) = 1/(ti+1)2 = (1i)2 [ 1 + t/(1-i)]-2
= i/2[ 1- 2t/(1-i) + 3t2/(1-i)24t3/(1-i)3 + ..]
2. Expand f(z) = (2z3
+1)/z2+z) about z = i
Answer: f(z) = (2z3
+1)/z2+z) = (2z-2) + (2z+1)/z(z+1)
(2i-2) +2(z-i) + 1/z + 1/z+1), Put zi = t
1/z = 1/(t+i) = 1/i(1+t/i)-1 = 1/i[1- t/I + t2/i3 - t3/i4 +..]
= -i + ( zi) +2 (-1)n(z - i)
n/t
n+1)
45 T, L series
1/(z+1) = 1/(t+i+1)
= 1/(i+1)[1+ t/(1+i)]-1
= 1/(1+i) [ 1- t/(1+i) + t2/(1+i)
2t3/(1+i)3 + -..
= (1-i)/2t/2i + t2/(1+i)3t3/(1+i)4+- ..
= - i/2(z - i)/2i +2 (-1)n(z-i)
n/(1+it)
n+1
Substituting, f(z) = (i/3 - 3/2) + (3 + i/2)(z-i) +
2 (-1)n [1/in+1 +1/(1+i)n+1 (z-i)n
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46 TL Series
2. Expand f(z) = 1/(z-1)(z-2) in the region
a. |z| < 1
Answer: Resolving into partial fractionsf(z) = 1/(z-1)(z-2) = 1/(z-2)1/(z-1)
= -1/2 (1- z/2)-1
+(1 - z)-1
For |z| < 1 both |z/2| and |z| are < 1
So -1/2 (1- z/2)-1
+(1 - z)-1
=
= -(1+z/2+z2/4 + z
3/8+.+(1+z+z2+z3+.)
= + z+ 7/8 z2
+ 15/16 z3 + .
47 TL Series
Expand f(z) = 1/(z-1)(z-2) in the region
for 1 < |z| < 2
Answer: Writing f(z) as - .1/(1-z/2)1/z(1-z-1)-1
Both |z/2| and |z-1
| are less than 1. hence
= - (1+z/2 +z2/4 + z
3/8 + .) 1/z(1+ z-1 + z-2+ z-3 + z-4 + ..)
= - z-4 - z-3 - z-2 - z-1 -1/4 z -1/8 z2- .
Which is a Laurent series
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48 TL Series
Expand f(z) = 1/(z-1)(z-2) in the region
|z| > 2
Answer: For |z| > 2,
f(z) = 1/z(1 - 2z-1
)1/z(1 - z-1)-1
= z-1
( 1 + 2z-1
+ 4z-2
+ 8z-3 + ..) -z-1(1 + z-1 + Z-2 + Z-3+..)
= .+ 8Z-4
+ 4Z-3
+ 2Z-2
- 1 - Z - Z2
49 TLSeries
Expand f(z) = 1/(z-1)(z-2) in the region
0< |z1 | < 1.
Answer: Writing f(z) = 1/(z-1)(z-2) = 1/[(z-1)-1]1/(z-1)
= -(z-1)-1[1-(z-1)]-1
= -(z-1)-1
- [ 1+ (z-1) +(z-1)2
+ (z-1)3 .. ]
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50 TL Series
5. Expand f(z) = z/(z-1)(z-3) in power series in
a. 1< |z | < 3.
Answer: Resolving into partial fractions
f(z) = z/(z-1)(z-3) = -1/2/(z-1)+ 3/2/(z-3)
= - 1/z(1-1/z) + 3/2 1/-3(1-z/3)
= -1/2z(1-1/z)-1 (1 - z/3)-1
= -1/2z( 1+1/z+ 1/z2 + +..) ( 1+ z/3 + z2/9 + z3/27+..)
51 CA
6. Expand f(z) = z/(z-1)(z-3) in power series in
|z1 | < 2.
Answer: f(z) = -1/2/(z-1)+ 3/2/(z-3),
Put z = u +1, = - . 1/u + 3/2 . 1/(u+1-3) =
= - 1/(z-1) + 3/2u[ 1 - 2/u]
-1
=
= - 1/(z-1) + 3/2(z-1) +3/(z-1)2
+ 6/(z-1)3 +
52 Singularities
Zero: The value of z for which f(z) = 0, is called zero of f(z)
A point is a singular point if it ceases to be
Analytic at that point. If f(z) is analytic everywhere except for the point z = a, then z = a is
Called isolated singularity.
In f(z)=1/(z-a)n
, z = a is called a pole of order n.
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53 CA
1. In (zsinz)/z2
, z = 0, is a removableSingularity, since
(zsinz)/z2 = 1/z2 [ z - {zz3/3! +z5/5! - + }]
i.e. no negative powers of z are in the expansion.
2. Poles of 1/(1 - ez) are the solutions of (1- e
z) = 0 or
z = 2ni, n = 0,1,2,3, ..
Residue: The coefficient of (z-a)-1
, in the expansion of
f(z) around an isolated singularity is the residue of f(z) at
that point.
54 Residue theorem
Residue f(a) = 1/2icf(z)dz
that iscf(z)dz = 2i f(a)
1. If f(z) has a simple pole at z = a,
then Resf(a) = Lt [ (z-a) f(z)
za2. If f(z) has a pole of order n at z = a, then
Res f(a) = 1/(n-1)! [ dn-1
/dzn-1
(z-a)n
f(z)] at z = a
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55 CA
Determine poles of f(z) = z
2
/(z-1)
2
(z+2) and residue at each pole, hence evaluatecf(z)dz
Where c is the circle |z| = 5/2
Answer: Limit as z -2, of (z+2)f(z) = 4/9
Simple pole at z = -2 and residue = 4/9
Limit z1, of (z-1)2 f(z), -1 is a pole of order 2,
Res f(1) = d/dz(z-1)2f(z) at z = 1, = 5/9
cf(z)dz = 2i(sum of the residues) = 2i(4/9+5/9)
56 CA-Residues
Evaluatec(z-3)/(z2 +2z +5)dz, where c is the circle |z|=1
Answer: Solving z2 +2z +5 = 0,
z = -1+2i, -1-2i. Both the
Poles of f(z) lie outside the circle |z| = 1,
hence (z-3)/(z2
+2z +5) is analytic in |z| = 1
and so by Cauchys theorem
c(z-3)/(z2 +2z +5)dz = 0.
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57 CA-Residues-Integrals
Evaluatec(z-3)/(z2
+2z +5)dz, where c is the circle
|z+1-i|=2.
Answer: the poles of (z-3)/(z2 +2z +5) are -1+2i
and -1-2i. The pole -1+2i lies inside the circle
|z + 1 - i| = 2. So f(z) is analytic except at this point
c(z-3)/(z2 +2z +5)dz = Resf(-1+2i).
So limit z -1+2i of [z-(-1+2i)f(z)] = +i
Hence by Residue theoremc(z-3)/(z2
+2z +5)dz
= 2i(i + ) =(i - 2)
58 CA-Residues-Integrals
Evaluatec(z-3)/(z2
+2z +5)dz, where c is the circle
|z + 1 + i| = 2.
Answer: the poles of (z-3)/(z2 +2z +5) are -1+2i
and -1-2i. The pole -1- 2i lies inside the circle
|z + 1 + i| = 2. So f(z) is analytic except at this point
c(z-3)/(z2
+ 2z +5)dz = Resf(-1 - 2i).
So limit z -1 - 2i of [z-(-1 + 2i)f(z)] = -i
Hence by Residue theoremc(z-3)/(z2 +2z +5)dz
= 2i(-i + ) =(i + 2)
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59 CA-Residues-Integrals
Evaluatec (tanz) dz, where c is the circle |z | = 2.
Answer: The poles of tanz = sinz/cosz are at
z = (2n+1)/2, n = 0,1,2,3. Of these only z = -/2,/2, lie inside the circle |z| = 2. So f(z) is analytic except
at this point, soc(tanz)dz = 2i[Resf(/2) + Resf(-/2)]
Res f(/2) = So limit z /2 sinz/-sinz = -1
Res f(-/2) = So limit z -/2 sinz/-sinz = -1
Hence by Residue theoremc(tanz)dz = 2i(-1 - 1) = -4i
60 CA-Residues-Integrals
Evaluatec (sinz2
+ cosz2)/(z-1)2 (z -2) dz, where
c is the circle |z| = 3.
Answer: The pole z = 1 is of second order and z = 1 is a
simple pole. So f(z) is analytic except at these points.
c
(sinz2 + cosz2)/(z-1)2 (z -2) dz = Sum of Residues at
1 and 2. Residue at z = 1,
= d/dz [(sinz2 + cosz2)/(z -2)]= 2 + 1
Residue at z = 2, Lt z 2, (sinz2 + cosz2)/(z-1)2 = 1
c (sinz2
+ cosz2)/(z-1)2 (z -2) dz = 2i(2 + 1 + 1) = 4( +1)i