Complex Analysis - Midterm examination - Solutions

Problem 1 : Let z0 =√

3− 3i.

1. Express z0 in polar form (in the form r exp(θi)).Answer: Let us first compute the modulus of z0:

|z0| =√√

32

+ (−3)2 =√

3 + 9 =√

12 = 2√

3.

Therefore, we can easily get the argument of z0:

Arg(z0) = Arg

√3− 3i

2√

3= Arg

(1

2−√

3

2i

)= −π

3

up to a multiple of 2π. Finally, the polar form of z0 is

z0 = 2√

3 exp(−π

3i).

2. Solve the equation z3 = z0.Answer: We know that the nonzero complex number z0 has exactly 3 cube roots.More precisely, they can be written

rk = 3√|z0| exp

(Arg(z0) + 2kπ

3i

)for k ∈ {0, 1, 2}. Finally, the three solutions of the equations are

r0 =6√

12 exp(−π

9i)

; r1 =6√

12 exp

(5π

9i

); r2 =

6√

12 exp

(11π

9i

).

Problem 2 : See other page.

Problem 3 : Solve the equations (find all solutions in C):

1. z2 − 2z + (1 + 2i) = 0.Answer: We can compute

z2 − 2z + (1 + 2i) = (z − 1)2 + 2i

so the solutions z of z2 − 2z + (1 + 2i) = 0 satisfy (z − 1)2 = −2i. We have

−2i = 2 exp(−π

2i)

so the square roots of −2i are

δ0 =√

2 exp(−π

4i)

=√

2(

cosπ

4− sin

π

4i)

= 1− i1

andδ1 = −δ0 = −1 + i.

Finally the solutions of the original equations satisfy z − 1 = δ0 or z − 1 = δ1 orequivalently, these solutions are

z0 = 1 + δ0 = 2− i and z1 = 1 + δ1 = i.

2. exp(2z) =√

3− i.Answer: Let us put

√3− i in polar form:

|√

3− i| =√√

32

+ (−1)2 =√

3 + 1 = 2

and √3− i2

= exp(−π

6i).

So√

3− i = 2 exp(−π

6i)

and we have to solve the equation

exp(2z) = 2 exp(−π

6i)

= exp(

log(2)− π

6i).

As exp is periodic of period 2πi, this equation is equivalent to

2z = log(2)− π

6i+ 2kπi

for k ∈ Z. Finally, the solutions are

zk =log(2)

2+(− π

12+ kπ

)i

for k ∈ Z.

Problem 4 : Compute the following limits when they exist. When they do not exist,prove it.

1. limz→1+i

z − 1− iz2 − 2i

.

Answer: For z ∈ C such that z2 6= 2i, we have

z − 1− iz2 − 2i

=z − 1− i

(z − (1 + i))(z + (1 + i))=

1

z + 1 + i

so

limz→1+i

z − 1− iz2 − 2i

= limz→1+i

1

z + 1 + i=

1

2 + 2i=

2− 2i

8=

1

4− 1

4i.

2. limz→1+i

Re(z − 1− i)z2 − 2i

.

Answer: Let f : C \ {1 + i,−1− i} → C be defined by

f(z) =Re(z − 1− i)

z2 − 2i=

Re(z − 1− i)(z − 1− i)(z + 1 + i)

.

2

Suppose thatlimz→1+i

f(z) = L ∈ C.

Let g : R→ C defined by g(t) = (1 + t) + i. As g(t) −−→t→0

1 + i, we have

L = limt→0

f(g(t)) = limt→0

Re(t)

t(2 + t+ 2i)= lim

t→0

t

t(2 + t+ 2i)

= limt→0

1

2 + t+ 2i=

1

2 + 2i.

On the other hand, let h : R→ C defined by h(t) = 1 + (t+ 1)i. As h(t) −−→t→0

1 + i,

we have

L = limt→0

f(h(t)) = limt→0

Re(ti)

ti(2 + (t+ 2)i)= lim

t→0

0

ti(2 + (t+ 2)i)

= limt→0

0 = 0.

This is a contradiction as the limit L should be unique. Finally, f does not convergeat 1 + i.

Problem 5 : For each of the following functions, tell if it is analytic and, if it is the case,compute its derivative. In both cases justify precisely your answer.

1. f(z) = z2 exp(z) for z ∈ C;Answer: Using the usual rule for differentiability of products, we see that f isanalytic and for z ∈ C,

f ′(z) =d z2 exp(z)

d z=

d z2

d zexp(z) + z2

d exp(z)

d z= (2z + z2) exp(z).

2. g(z) =1

zfor z ∈ C∗;

Answer: Let us rewrite this function as a function from R2 \ {0} to R2. As we haveg(z) = 1/z = z/|z|2, we get

gR(x, y) =

(x

x2 + y2,

y

x2 + y2

).

We deduce that gR is differentiable on R2 \ {0} and

∂gR,1(x, y)

∂x=x2 + y2 − 2x2

(x2 + y2)2=

y2 − x2

(x2 + y2)2

and∂gR,2(x, y)

∂y=x2 + y2 − 2y2

(x2 + y2)2=

x2 − y2

(x2 + y2)2

so g is not analytic over C as it does not satisfy the Cauchy-Riemann equation

∂gR,1(x, y)

∂x=∂gR,2(x, y)

∂y.

3

3. h(z) = log√x2 + y2− i tan−1

x

yfor z = x+ yi satisfying x, y ∈ R and y > 0. Here,

log and tan−1 are the usual real functions. We recall thatd tan−1(t)

d t=

1

1 + t2.

Answer: Let us rewrite this function as a function from {(x, y) ∈ R2 | y > 0} to R2:

hR(x, y) =

(log√x2 + y2,− tan−1

x

y

).

We deduce that

∂hR,1(x, y)

∂x=

1√x2 + y2

∂√x2 + y2

∂x=

1√x2 + y2

x√x2 + y2

=x

x2 + y2;

∂hR,1(x, y)

∂y=

1√x2 + y2

∂√x2 + y2

∂y=

1√x2 + y2

y√x2 + y2

=y

x2 + y2;

∂hR,2(x, y)

∂x= − 1

1 + (x/y)2∂(x/y)

∂x= − y2

x2 + y21

y= − y

x2 + y2;

∂hR,2(x, y)

∂y= − 1

1 + (x/y)2∂(x/y)

∂y=

y2

x2 + y2x

y2=

x

x2 + y2.

so h is analytic on its domain as hR is differentiable on {(x, y) ∈ R2 | y > 0} (itspartial derivatives are continuous) and satisfies the Cauchy-Riemann equations

∂hR,1(x, y)

∂x=∂hR,2(x, y)

∂y∂hR,2(x, y)

∂x= −∂hR,1(x, y)

∂y.

Moreover, we know in this case that

h′(z) = h′(x+ yi) =∂hR,1(x, y)

∂x+∂hR,2(x, y)

∂xi =

x− yix2 + y2

=1

z.

4

Complex Analysis - Mid-term exam - Solutions

Problem 2

The number z ∈ C is represented in the following diagram. Construct the pointrepresenting

3

2z exp

(−π

4i)

+ i+ 1

on the diagram. Justify the construction geometrically by drawing some other points andmarking clearly the steps of the construction.

0 1

i

z

32z3

2z exp

(−π

4i)

32z exp

(−π

4i)

32z exp

(−π

4i)

+ i+ 1

−π4

i+ 1

5