Complex variables: Exam 1 Solutions 7/9/9

Question 1

Determine the following limits, or explain why the limit in question does notexist.

• limz→1+i

(z4 + 2iz2 + 8

z2 − 3iz − 3 + i

)When z = 1 + i, we have z2 = 1 − 1 + 2i = 2i, so z4 = −4,so z4 + 2iz2 + 8 = −4 + 2i(2i) + 8 = 0 and z2 − 3iz − 3 + i =2i− 3i(1 + i)− 3 + i = 0.So we may use L’Hopital’s rule, giving:

limz→1+i

(z4 + 2iz2 + 8

z2 − 3iz − 3 + i

)= lim

z→1+i

(4z3 + 4iz

2z − 3i

)

= 4 limz→1+i

(z(z2 + i

2z − 3i

)= 4

((1 + i)(2i+ i

2(1 + i)− 3i

)= 12

(i− 1

2− i

)=

(12

5

)(i− 1)(2 + i) =

(12

5

)(−3 + i) = −7.2 + 2.4i.

Alternatively, we can factor:

z4 + 2iz2 + 8 = (z2 + 4i)(z2 − 2i) = (z2 + 4i)(z − 1− i)(z + 1 + i),

z2 − 3iz − 3 + i = (z − 1− i)(z + 1− 2i),

limz→1+i

(z4 + 2iz2 + 8

z2 − 3iz − 3 + i

)= lim

z→1+i

(4z3 + 4iz

2z − 3i

)= lim

z→1+i

((z2 + 4i)(z − 1− i)(z + 1 + i)

(z − 1− i)(z + 1− 2i)

)= lim

z→1+i

((z2 + 4i)(z + 1 + i)

(z + 1− 2i)

)=

((2i+ 4i)(2 + 2i)

(2− i)

)= 12

((i− 1)(2 + i)

5

)=

(12

5

)(−3+i) = −7.2+2.4i.

• limz→0

(ln(1− z2)

sin2(2z)

)Since sin(0) = ln(1) = 0, this is a standard ”0/0” limit, so we mayapply L’Hopital’s rule, giving, by the chain rule:

limz→0

(ln(1− z2)

sin2(2z)

)= lim

z→0

(−(

2z1−z2

)4 sin(2z) cos(2z)

)

= limz→0

(−(

2z1−0

)4 sin(2z) cos(0)

)= −

(1

2

)limz→0

(z

sin(2z)

)Here we again have a standard ”0/0” limit, so we may apply L’Hopital’srule, giving:

−(

1

2

)limz→0

(z

sin(2z)

)= −

(1

2

)limz→0

(1

2 cos(2z)

)= −

(1

2

)(1

2 cos(0)

)= −1

4.

So the required limit exists and is −1

4.

• limz→0

(<(z2)=(z)

z3

)Put z = reiθ, where r > 0 and θ are both real.Then we have z3 = r3e3iθ. Also <(z2) = <(r2e2iθ) = r2 cos(2θ) and=(z) = =(reiθ) = r sin(θ), so the required limit becomes:

limz→0

(<(z2)=(z)

z3

)= lim

r→0+

(r3 cos(2θ) sin(θ)

r3e3iθ

)= lim

r→0+

(e−3iθ cos(2θ) sin(θ)

).

If we approach z = 0 along the positive x-axis, we have θ = 0, so thelimit is e0 cos(0) sin(0) = 0.

If we approach z = 0 along the line θ =π

6, the limit is e

iπ2 cos

(π3

)sin(π

6

)= − i

4.

Since these two limits differ, the required limit does not exist.

Alternatively, we could first take z 6= 0 real, giving limit zero, since=(z) = 0.Then we take z = ti, where t 6= 0 is real, so we have z3 = −it3, =(z) = t

and <(z2) = <(−t2) = −t2, giving the limit limt→0

−t2(t)−it3

= −i.Again, since these two limits differ, the required limit does not exist.

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Question 2

Sketch the following sets in the complex plane and for each identify whetherthe set is open, closed or neither and whether or not the set is bounded,connected or compact.For each of these sets also give a parametrization or parametrizations of itsboundary, as appropriate, where the boundary is traced counter-clockwisewith respect to an observer in the set.

• A = {z : 4<(z) ≤ =(z)− 4}.

Put z = x+ iy, with x and y real.Then we need 4x ≤ y − 4, or y ≥ 4x+ 4.So the region A is the closed-half-plane lying on or above the straightline y = 4x+ 4, which has slope 4 and intecepts (0, 4) and (−1, 0).Then A is closed, unbounded, non-compact and connected.Its boundary is the line y = 4x+ 4, which is parametrized as x = t andy = 4t+ 4, for any real t.The line goes up from bottom left to top right.

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4

• B = {z : |z − 2| ≤ 2 and |z − 2i| ≤ 2}.

The region B is the closed region common to the two circles of ra-dius 2, one centered at (2, 0), the other at (0, 2).It is symmetrical about the line y = x and is bounded by two circulararcs, which meet at the points z = 0 = (0, 0) and z = 2 + 2i = (2, 2)on that line.Then B is closed, bounded, connected and compact.

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The part of the boundary on the circle |z−2| = 2 goes counter-clockwise

from z = 2 + 2i to z = 0, so is z = 2 + 2eis,π

2≤ s ≤ π.

The part of the boundary on the circle |z − 2i| = 2 goes counter-clockwise from z = 0 to z = 2 + 2i, so is z = 2i+ 2eit, −π ≤ t ≤ 0.

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• C =

{z = reiθ : 0 < r ≤ 3 and − 2π

3≤ θ ≤ 0

}.

This is a sector of the disc centered at the origin, of radius 3 units,in the third and fourth quadrants bounded above by the positive x-axis and by the line through the origin sloping up at 60 degrees, exceptthat the origin is deleted.

Since the origin is deleted, C is neither closed nor open, is bounded,non-compact and connected. The boundary of C has three pieces:

– The circular arc from z = 3e−2iπ3 to z = 3:

z = 3eiθ, −2π

3≤ θ ≤ 0,

– The real axis, from z = 3 to z = 0

z = 3− s, 0 ≤ s ≤ 3,

– The line of slope 60 degrees through the origin from z = 0 to

z = 3e−2iπ3 :

z = te−2iπ3 , 0 ≤ t ≤ 3.

Note that the boundary of C includes the origin, even though it is notin C, because it is a limit point of C.

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Question 3

Let g(z) =2z − 1

z + 2be defined for any complex number z 6= −2.

• Prove that the function g(z) is analytic on its domain and compute itsderivative from first principles.We need to show that the complex derivative of g exists everywhere,so we need to compute the limit, for z 6= −2 and w 6= −2:

g′(z) = limw→z

g(w)− g(z)

w − z

= limw→z

2w−1w+2− 2z−1

z+2

w − z

= limw→z

(2w − 1)(z + 2)− (2z − 1)(w + 2)

(w − z)(z + 2)(w + 2)

=1

(z + 2)2limw→z

2wz − z + 4w − 2− (2zw − w + 4z − 2)

(w − z)

=1

(z + 2)2limw→z

5w − 5z

(w − z)=

1

(z + 2)2limw→z

5 =5

(z + 2)2.

So g is analytic on its domain, as required.

• Is <(g) harmonic on its domain? Explain.Yes every analytic function obeys the Cauchy-Riemann equations andits real and imaginary parts are then harmonic.

• Find, at least locally, a function f(z) such that d(f(z)) = g(z)dz.For x real, we compute the integral:∫ (

2x− 1

x+ 2

)dx =

∫ (2u− 5

u

)du =

∫ (2− 5

u

)du

= 2u− 5 ln(u) = 2(x+ 2)− 5 ln(x+ 2).

Here we put x = u− 2, u = x+ 2, du = dx.So f(z) = 2(z + 2)− 5 ln(z + 2) will do.

• Does the domain of your solution for f(z) match the domain of thefunction g(z)?Explain.No. The function f(z) requires a branch cut from −2 to infinity,whereas the function g(z) does not.

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Question 4

Let u = x4 + y4 − 6x2y2 + 2x2 − 2y2 + 1

• Prove that u is harmonic and find its harmonic conjugate v.

We have:x2 − y2 = <(z2),

x4 + y4 − 6x2y2 = <(z4).

So u = <(z4 + 2z2 + 1) = <((z2 + 1)2).So u is harmonic, with harmonic conjugate:

v = =(z4 + 2z2 + 1) = 4x3y − 4xy3 + 4xy + c.

Here c is an arbitrary real constant.

Alternatively, we solve the Cauchy-Riemann equations:

vx = −uy = −4y3 + 12x2y + 4y,

v = −4xy3 + 4x3y + 4xy + k(y),

0 = vy − ux = k′(y)− 12xy2 + 4x3 + 4x− 4x3 + 12xy2 − 4x = k′(y).

So k is constant and the harmonic conjugates of u are the functionsv = −4xy3 + 4x3y + 4xy + k, where k is an arbitrary constant.

• Express f = u+ iv as a function of z = x+ iy and show that:

f ′(z) = ∂xf = −i∂yf.

Explain why f is unique, given that f(0) = 1.The freedom in choosing the harmonic conjugate v of an harmonic func-tion u is just a constant, since the derivatives vx and vy are known.The constant is fixed uniquely by specifying the value of the harmonicconjugate at one point.Here the requirement f(0) = 1 gives v(0) = 0, so v is uniquely deter-mined, as required.

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We have u+ iv = f(z), where f(z) = z4 + 2z2 + 1 = (z2 + 1)2.

Then we have, by the chain rule:

f ′(z) = 2(z2 + 1)(2z) = 4z(z2 + 1) = 4(x+ iy)(x2 − y2 + 1 + 2ixy)

= 4x3 − 12xy2 + 4x+ iy(12x2 − 4y2 + 4)

Also we have:

fx = ux + ivx = 4x3 − 12xy2 + 4x+ i(12x2y − 4y3 + 4y)

= 4x3 − 12xy2 + 4x+ iy(12x2 − 4y2 + 4) = f ′(z),

−ify = vy−iuy+vy = 4x3−12xy2+4x−i(4y3−12x2y−4y) = fx = f ′(z).

• When f(0) = 1, factorize the function u2 + v2 as a product of complexfunctions, each linear in the variables x and y (or, equivalently, linearin the variables z and z).

We have:

u2 + v2 = (u+ iv)(u− iv) = (z2 + 1)2(z2 − 1)2

= (z + i)(z + i)(z − i)(z − i)(z + i)(z + i)(z − i)(z − i)

= (x+ i(y + 1))(x+ i(y + 1))(x+ i(y − 1))(x+ i(y − 1))

(x− i(y − 1))(x− i(y − 1))(x− i(y + 1))(x− i(y + 1)).

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Question 5

Consider the transformation T : z → (1 + i)z+ 3− 4i, defined for any z ∈ C.

• Find the image A′B′C ′D′ under T of the square ABCD with verticesA = 1 + i, B = −1 + i, C = −1− i and D = 1− i.How do the areas of ABCD and A′B′C ′D′ compare?

We have:

– A′ = T (A) = (1 + i)(1 + i) + 3− 4i = 2i+ 3− 4i = 3− 2i,

– B′ = T (B) = (1 + i)(−1 + i) + 3− 4i = −2 + 3− 4i = 1− 4i,

– C ′ = T (C) = −(1 + i)(1 + i) + 3− 4i = −2i+ 3− 4i = 3− 6i,

– D′ = T (D) = −(1 + i)(−1 + i) + 3− 4i = 2 + 3− 4i = 5− 4i.

Plotting we see that A′B′C ′D′ is a square of side length:

|A′B′| = |B′ − A′| = | − 2− 2i| =√

4 + 4 = 2√

2.

So the area of A′B′C ′D′ =(

2√

2)2

= 8 units of area.

This is double the area of ABCD which is a square of side length 2, sohas area 4 units of area.

• Find the fixed points of the transformation T , if any.

The fixed points are the solutions of the equation:

T (z) = z,

(1 + i)z + 3− 4i = z,

iz = 4i− 3,

z = −i(4i− 3) = 4 + 3i.

Check:

T (4 + 3i) = (1 + i)(4 + 3i) + 3− 4i = 4− 3 + 4i+ 3i+ 3− 4i = 4 + 3i.

So the only fixed point is the point z = 4 + 3i = (4, 3).

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• Describe the transformation T geometrically.

We see that the transformation is a combination of a counter-clockwiserotation through forty-five degrees and a dilation by a factor of

√2

centered at the point (4, 3).

In fact, since 1 + i =√

2 eiπ4 , we have:

T =√

2 eiπ4 z + 3− 4i.

The factor of√

2 gives the dilation.

The factor of eiπ4 gives the rotation.

The term 3− 4i gives a translation, which has the effect of moving thecenter of the transformation to the invariant point (4, 3).

• Find a formula for the inverse transformation, T −1.We solve the equation:

w = T (z) = (1 + i)z + 3− 4i,

(1− i)w = (1− i)(1 + i)z + (1− i)(3− 4i) = 2z − 1− 7i,

z =1

2((1− i)w + 1 + 7i).

So the inverse transformation is:

T −1(z) =1

2((1− i)z + 1 + 7i).

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• Find the image under the transformation T of the circle |z| = 5 andsketch the circle and its image on the complex plane.Also give an equation for the image circle.

Since the transformation is a combination of rotations, translationsand dilations, it maps straight lines to straight lines and circles to cir-cles and scales all distances by the dilation factor

√2.

So the image of the given circle which has radius 5 is a circle whoseradius is 5

√2.

Since the given circle has center at the origin, the image circle has itscenter at the image of the origin, which is the point T (0) = 3− 4i.So the image circle is the circle center 3− 4i = (3, 4) and radius 5

√2.

Its equation is |z − 3 + 4i| = 5√

2, or squaring:

50 = (x− 3)2 + (y + 4)2,

x2 + y2 − 6x+ 8y = 25.

Alternatively, we parametrize the given circle:

z = 5eit, t real.

Then the image of the given circle under the transformation T is thecollection of points:

z = T (5eit) = 5(1 + i)eit + 3− 4i.

Then we have:z − 3 + 4i = 5(1 + i)eit.

Taking the absolute value of both sides, we eliminate t, since |eit| = 1,for t real.So we get the equation of the image as:

|z − 3 + 4i| = |5(1 + i)eit| = 5|1 + i| = 5√

2.

This gives the same equation as before.Note that the fixed point of T lies on both circles.

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Question 6

Consider the function f(z) = (z − 4)12 , where f(8) = 2.

If the branch cut for f is the interval on the real axis (−∞, 4], determine thevalue of the following, explaining your results:

• f(4i)

• limt→0+ f(ti)

• limt→0− f(t(1 + i))

If instead the branch cut is along the line z = 4 + si, for s real and non-negative, which, if any, of these values changes?Explain your answer.

We write z − 4 = seiα, where s > 0 is the distance of the point z fromthe point 4 and α radians is the angle that the vector from 4 to z makes withthe positive x-axis.

Then f(z) = (z − 4)12 =√s e

iα2 .

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With the cut along the real axis on the interval (−∞, 4], we may take α inthe range −π < α < π, since then we have for the point z = 8, s = |z − 4| =|8− 4| = 4 and α = 0 and then f(8) =

√4 e0 = 2, as required.

Then we have:

• For f(4i), we have z − 4 = 4i− 4, so s =√

16 + 16 = 252 .

Also α =3π

4, so we have:

f(4i) = 254 e

3iπ8 = 2

54

(cos

(3iπ

8

)+ i sin

(3iπ

8

))= 0.91018+2.19737i.

Alternatively, we algebraically solve the equation:

z2 = 4i− 4 = x2 − y2 + 2ixy, z = x+ iy,

x2 − y2 = −4, xy = 2,

|z|2 =∣∣z2∣∣ = |4i− 4| =

√32 = x2 + y2,

x2 =1

2(√

32− 4) = 2(√

2− 1), y2 =1

2(√

32 + 4) = 2(√

2 + 1).

The desired solution lies in the first quadrant, since the angleα

2=

3π

8radians is an acute angle. So we get:

f(4i) =√

2

(√√2− 1 + i

√√2 + 1

)= 0.91018 + 2.19737i.

• limt→0+ f(ti)Here we see that s = limt→0+ |ti− 4| = 4.Also since we are approaching the origin from above, we see that theargument of z − 4 goes to π, so in the limit, we have α = π.So we get:

limt→0+

f(ti) =√

4eiπ2 = 2i.

• limt→0− f(t(1 + i))Here we see that s = limt→0− |t(1 + i)− 4| = 4.Also since we are approaching the origin from below, we see that theargument of z − 4 goes to −π, so in the limit, we have α = −π.So we get:

limt→0+

f(ti) =√

4e−iπ2 = −2i.

Note that in this case the function cannot be continuous across the cut.

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With the cut along the line z = 4 + si, for s real and non-negative, we may

take α in the range −3π

2< α <

π

2, since then we have for the point z = 8,

s = |z − 4| = |8− 4| = 4 and α = 0 and then f(8) =√

4 e0 = 2, as required.Then we have:

• For f(4i), we have z − 4 = 4i− 4, so s =√

16 + 16 = 252 .

Also α = −5π

4, so we have:

f(4i) = 254 e

−5iπ8 .

This lies in the third quadrant and is the negative of our previoussolution, so we now have:

f(4i) = −√

2

(√√2− 1 + i

√√2 + 1

)= −0.91018− 2.19737i.

• limt→0+ f(ti)Here we see that s = limt→0+ |ti− 4| = 4.Also since we are approaching the origin from above, but to the left ofthe cut line, we see that the argument of z − 4 goes to −π, so in thelimit, we have α = −π.So we get:

limt→0+

f(ti) =√

4e−iπ2 = −2i.

• limt→0− f(t(1 + i))Here we see that s = limt→0− |t(1 + i)− 4| = 4.Also since we are approaching the origin from below, we see that theargument of z − 4 goes to −π, so in the limit, we have α = −π.So we get:

limt→0+

f(ti) =√

4e−iπ2 = −2i.

Note that in this case f is analytic near the origin.

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Question 7

Write the function g =4x− 8y

x2 + y2as the real part of an analytic function.

Hint: consider the function z−1.Hence, or otherwise, solve the equation:(

∂2x + ∂2

y

)f = g.

We have, with z = x+ iy:1

z=

x− iyx2 + y2

,

i

z=

y + ix

x2 + y2,

4− 8i

z=

4x− 8y + i(−4y − 8x)

x2 + y2,

So if we write: G(z) = (4 − 8i)z−1, then G is analytic and g = <(G), asrequired.To solve the equation:

(∂2x + ∂2

y

)f = g, we first solve the equation

(∂2x + ∂2

y

)F =

G, and then f = <(F ).So we solve:

4∂z∂zF =4− 8i

z,

∂z∂zF =1− 2i

z,

Integrating both sides of this equation with respect to z (ignoring integrationconstants), treating z as a constant, gives:

∂zF = (1− 2i) ln(z).

Integrating both sides of this equation with respect to z (again ignoringintegration constants), treating z as a constant, gives the solution:

F = (1− 2i)z ln(z).

The general local solution is then F = (1− 2i)z ln(z) + H(z) + J(z), whereH is analytic and J is anti-analytic.So the required solution is:

f = <((1− 2i)z ln(z) +m(z)).

Here m(z) is an arbitrary analytic function.

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To write out the function f explicitly, in terms of x and y, we need:

<((1− 2i)z ln(z)) = <((1− 2i)(x− iy)(ln(r) + iθ))

= <((x− 2y − i(2x+ y))(ln(r) + iθ))

= (x− 2y) ln(r) + (2x+ y)θ

=1

2(x− 2y) ln(x2 + y2) + (2x+ y) arctan

(yx

).

Here we have written z = reiθ with r =√x2 + y2 > 0 and θ = arctan

(yx

).

Also we have used the fact that ln(r) =1

2ln(r2)

=1

2ln(x2 + y2

).

So the required local solution is:

f =1

2(x− 2y) ln(x2 + y2) + (2x+ y) arctan

(yx

)+M(x, y).

Here M is an arbitrary harmonic function, or, equivalently, the real part ofan analytic function.

A somewhat simpler solution is to replace the (1 − 2i)z ln(z) term of the

solution by the term (1− 2i)z ln(zz

), which is legitimate, because the extra

term, which is −(1− 2i)z ln(z), is anti-analytic so is harmonic.

Then, compared to the above calculation, the θ term doubles and the rterm goes away, so now we have:

f = <(

(1− 2i)z ln(zz

)+ n(z)

)= 2(2x+ y) arctan

(yx

)+N(x, y).

Here n is analytic and N is harmonic.

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