Complex Analysis (MAT3705), University of South Africa (UNISA), 2024–2026, complete exam solution pack

Memorandum MAT3705 Nov 2015

QUESTION 1
1.1

z−3 (z − 3)(z + 3)
( ) =
z+3 (z + 3)(z + 3)
(z − 3)(z + 3)
= 2
|z + 3|
zz + 3z − 3z − 9
= 2
|z + 3|
2
|z| − 9 3(z − z)
= 2 + 2
|z + 3| |z + 3|

Since z − z is purely imaginary (take z = x + iy then z − z = 2iy) we have that
2
z−3 |z| − 9
Re( )= 2
z+3 |z + 3|

and if
2
|z| − 9
2 = 0 then |z| = 3.
|z + 3|
1.2
Suppose that z = x + iy where x, y ∈ R
Then
z+3
< 1 ⇔ |z + 3| < |z − i|
z−i
⇔ |(x + 3) + iy| < |x + (y − 1)i|
p p
⇔ (x + 3)2 + y 2 < x2 + (y − 1)2
⇔ (x + 3)2 + y 2 < x2 + (y − 1)2
⇔ 6x + 9 < −2y + 1
⇔ −2y > 6x + 8
⇔ y < −3x − 4



So the region is {x + iy | y < −3x − 4 } which is the region below the straight line y < −3x − 4
The set is open since it contains none of its boundary points.

QUESTION 2
2.1


1 z
sinh z = 3i ⇐⇒(e − e−z ) = 3i ⇐⇒ ez − e−z = 6i
2
⇔ (ez )2 − 6i(ez ) − 1 = 0

Substitute w = ez to obtain the equation w2 − 6iw − 1 = 0.



1

,Then
√
6i ± −36 + 4
w =
√2
6i ± 32i2
=
2√
= 3i ± 2 2i
√
= (3 ± 2 2)i




√ √ π
ex+iy = (3 ± 2 2)i = (3 ± 2 2)ei 2
√
⇔ x = ln(3 ± 2 2)
and
π
y = + 2nπ
2
so
√ π
z = ln(3 ± 2 2) + i( + 2nπ) n ∈ Z
2
or sinh z = sinh x cos y + i cosh x sin y = 3i ⇔ sinh x cos y = 0 and cosh x sin y = 3
cos y = 0 ⇔ y = π2 .
π 1 x −x x 2 x x
√
6± 36−4
√
Then cosh
√ x sin 2 = 3 ⇔ 2 (e + e ) = 3 ⇔ (e ) − 6e + 1 = 0 ⇔ e = 2 = 3±2 2 ⇔ x =
ln(3 ± 2 2) √
Then z = ln(3 ± 2 2) + i( π2 + 2nπ) n ∈ Z.
2.2 For z = x + iy where x, y ∈ R we have

eiz = ei(x+iy) = e−y+ix = e−y eix = e−y (cos x + i sin x)

Hence
Im eiz = e−y sin x


and

Im eiz = e−y sin x = 0
1
( y is never zero)
e
⇔ sin x = 0 and y ∈ R
⇔ x = nπ, y ∈ R
Hence
z = nπ + iy, n ∈ Z, y ∈ R

.
QUESTION 3.

f (z) = (z + 1)3 − 3z = (x + 1)3 − 3(x + 1)y 2 − 3x + i(y 3 + 3y − 3(x + 1)2 y)

So

u(x, y) = (x + 1)3 − 3(x + 1)y 2 − 3x
v(x, y) = y 3 + 3y − 3(x + 1)2 y


2

, ux = 3(x + 1)2 − 3y 2 − 3 vx = −6(x + 1)y
uy = −6(x + 1)y vy == 3y 2 + 3 − 3(x + 1)2
We see that ux = −vy rather than ux = vy and also uy = vx rather than uy = −vx .
This means that uy = −vx can only be valid when −6(x + 1)y = 0 i.e. at x = −1 or y = 0
For the case x = −1 then ux = −3y 2 − 3 = 3y 2 + 3 = vy ⇔ 6(y 2 + 1) = 0 ⇔ y = ±i.
Similarly if y = 0 then ux = 3(x + 1)2 − 3 = 3 − 3(x + 1)2 = vy ⇔ 6(x + 1)2 − 6 = 0 ⇔ 6[(x + 1) − 1][(x +
1) + 1] = 0 ⇔ x = 0 or x = −2.
So the function is differentiable only at the points (−1, i), (−1, i), (0, 0), (−2, 0)
But any neighbourhood of these points contains point where f is not differentaible so f (z) is nowhere
analytic.
QUESTION 4.
4.1
eiz
Z
2
dz
|z−i|=3 z + 9

The integrand has singularities at z = ±3i. Since |3i − i| < 3 but |−3i − i| > 3 only z = 3i is inside
|z − i| = 3
hence

eiz eiz /(z + 3i)
Z Z
2
dz = dz
|z−i|=3 z + 9 |z−i|=3 (z − 3i)
= 2πi g(3i) where g(z) = eiz /(z + 3i)
ei(3i)
= 2πi
6i
π −3
= e
3
4.2 Z
tan z/2 dz
|z|=2 (z − π/2)3
sin z/2
The integrand has a singularities at π2 and also since tan(z/2) = cos z/2 where cos z/2 = 0 i.e.where
π
z/2 = 2 + kπ i.e.z = π + 2kπ, k ∈ Z.
But for all k ∈ Z |π + 2kπ| ≥ π > 2(not inside |z| = 2 ) and so z = π2 is the only singularity which lies in
the interior of the simple closed contour |z| = 2 (it is a pole of order 3)
Z
tan z/2 dz 2πi d d
= ( tan z/2)) |z=π/2
|z|=2 (z − π/2)3 2! dz dz
d sec2 (z/2)
= πi ( ) |z=π/2
dz 2
2 1
= πi( sec (z/2) sec (z/2) tan(z/2). ) |z=π/2
2 2
πi
= sec2 (z/2) tan(z/2) |z=π/2
2
πi
= sec2 (π/4) tan(π/4 )
2
πi √ 2
= .( 2) .1
2
= πi

QUESTION 5.




3

, 1 1 1
f (z) = + +
z+1 z+3 z−4
in the region 5 < |z − 4| < 7.
Here we may set
1 1
=
z+1 (z − 4) + 5
5
So for 5 < |z − 4| ( |z−4| < 1) we get

1 1 1 1 1 5 5 2
= = 5 = (1 − ( )+( ) − ...)
z+1 (z − 4) + 5 (z − 4) (1 + (z−4) ) (z − 4) z − 4 z − 4


Similarly for |z − 4| < 7( |z−4|
7 < 1) we get

1 1 1 1
= = ( )
z+3 (z − 4) + 7 7 1 + z−4
7
1 z−4 z−4 2 z−4 3
= (1 − ( )+( ) −( ) + ...)
7 7 7 7
So for 5 < |z − 4| < 7

1 1 1
f (z) = + +
z+1 z+3 z−4
52 5 2 1 (z − 4) (z − 4)2
= (... + − + + − ) + − ...)
(z − 4)3 (z − 4)2 z−4 7 72 73


QUESTION 6.
6.1
z
f (z) =
(z 2 + 1)(z 2 + 2z + 2)
has singularities
p where the denominator is zero i.e.where z 2 + 1 = 0 i.e.z = ±i and z 2 + 2z + 2 = 0 i.e.
z = (−2 ± 4 − 4(2))/2 = −1 ± i.
Of these only i and −1 + i are in the upper half palne. Both are simple poles.

z
(z 2 +2z+2) z
Res f (z) |z=i = d
|z=i or |z=i
2
dz (z + 1)
(z 2 + 2z + 2)(z + i)
i
=
2i(−1 + 2i + 2)
1 1 − 2i
= =
2(1 + 2i) 10



Res f (z) |z=−1+i
z
(z 2 +1)
= d 2
|z=−1+i
dz (z + 2z + 2)
z
(z 2 +1)
= d 2
|z=−1+i
dz (z + 2z + 2)
z
= |z=−1+i
2(z 2 + 1)(z + 1)



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