Complex Analysis:

Solution of Homework 1

Problem 1
Use the mathematical induction to verify the binomial formula
n  
X n
(z1 + z2 )n = z1n−k z2k ,
k
k=0

where n = 1, 2, 3, · · ·

solution:
First, we should check that the steps of n = 1 and n = 2 are right:

For n = 1:
   
1 1
(z1 + z2 ) = 1 × z1 + 1 × z2 = z1 + z2 (n = 1 is right!)
0 1

For n = 2:

(z1 + z2 )2 = z12 + 2z1 z2 + z22

= 1 × z12 + 2 × z1 z2 + 1 × z22
     
2 2 2 2
= z1 + z1 z2 + z22 (n = 2 is right!)
0 1 2

Now we can assume n = l is also right i.e.

l  
X l
l
(z1 + z2 ) = z1l−k z2k .
k
k=0

And then we use this relation to prove that n = l + 1 is also right:

l  
X l
(z1 + z2 )l+1 l
= (z1 + z2 ) × (z1 + z2 ) = (z1 + z2 ) × z1l−k z2k
k
k=0
l   l  
X l X l
= z1l−k+1 z2k + z1l−k z2k+1
k k
k=0 k=0
l   l−1  
X l X l
= z1l+1 + z1l−k+1 z2k + z1l−k z2k+1 + z2l+1
k k
k=1 k=0
l   l  
X l X l
= z1l+1 + z1l−k+1 z2k + z1l−k+1 z2k + z2l+1
k k−1
k=1 k=1
l    
l+1
X l l
= z1 + + z1l−k+1 z2k + z2l+1 .
k k−1
k=1

1
Recall the Pascal’s formula:
     
l l l+1
+ = .
k k−1 k

Therefore, we show that the n = l + 1 step is right, too:

l    
X l l
(z1 + z2 )l+1 = z1l+1 + + z1l−k+1 z2k + z2l+1
k k−1
k=1
l  
X l+1 (l+1)−k k
= z1l+1 + z1 z2 + z2l+1
k
k=1
l+1
X l + 1  (l+1)−k

= z1 z2k (n = l + 1 is right!)
k
k=0

By mathematical induction, we obtain binomial formula:

n  
X n
n
(z1 + z2 ) = z1n−k z2k for n = 1, 2, 3, · · ·
k
k=0

Problem 2
If f (x) is a polynomial with real coefficients, and a + ib is a root of f (x) with a and b being real numbers,
then its complex conjugate a − ib is also a root of f (x).

solution:
Because f (x) is a polynomial with real coefficients, we can express it as follows:
X
f (x) = cn xn where cn ∈ R
n

a + ib is a root of f (x) means:

X
f (a + ib) = cn (a + ib)n = 0.
n

Take complex conjugate on above equality:

X
f (a + ib) = cn (a + ib)n = 0 = 0
n
X X
= cn (a + ib)n = cn × (a + ib)n
n n
X
n
= cn × (a + ib) (∵ cn ∈ R)
n
X
= cn × (a − ib)n
n
X
= cn (a − ib)n
n

= f (a − ib)

Obviously, a − ib is also a root of f (x):

f (a − ib) = f (a + ib) = 0 = 0

2
Problem 3
In each case, sketch the set of points determined by the given condition:
(a) |z − 1 + i| = 1; (b) |z + i| ≤ 3; (c) |z − 4i| ≥ 4.

solution:
(a) |z − 1 + i| = 1 ≡ |z − (1 − i)| = 1

Figure 1

(b) |z + i| ≤ 3 ≡ |z − (−i)| ≤ 3

Figure 2

(c) |z − 4i| ≥ 4

3
 Figure 3

Problem 4
Sketch the set of points determined by the given conditions:
(a) Re (z̄ − i) = 2; (b) |2z̄ + i| = 4.

solution:
(a) Re (z̄ − i) = 2 ≡ Re (x − iy − i) = Re [x − i(y + 1)] = 2
⇒ x = 2 and y ∈ R.

Figure 4

(b) |2z̄ + i| = 4 ≡ |z̄ + 2i | = 2

⇒ |x − i(y − 12 )| = 2
⇒ (x, y) ∈ R2 and x2 + (y − 12 )2 = 22 .

Figure 5

4
Problem 5
Show that the hyperbola x2 − y 2 = 1 can be written as z 2 + z̄ 2 = 2.

solution:
Set x = Re (z) and y = Im (z); the hyperbola x2 − y 2 = 1 becomes:

x2 − y 2 = 1
2 2
≡ [Re (z)] − [Im (z)] = 1
 2  2
z + z̄ z − z̄
≡ − =1
2 2i
z 2 + z̄ 2 + 2z z̄ z 2 + z̄ 2 − 2z z̄
= + =1
4 4
z 2 + z̄ 2
= =1
2
≡ z 2 + z̄ 2 = 2

Problem 6
Which sets in the following sets are bounded?
(a) |z − 2 + i| ≤ 1; (b) |2z + 3| > 4; (c) Im z > 1; (d) Im z = 1; (e) 0 ≤ arg z ≤ π/4 (z ̸= 0); (f) |z − 4| ≥ |z|.

solution:
Recall the definition of bounded set: a set S is bounded if every point in S lies inside some circle |z| = R
(R is a finite number); otherwise, it is unbounded. Now let us show the plot of above sets:

Figure 6

Obviously, only the set (a) is bounded.

5
Problem 7
Use de Moivre’s formula to derive the following trigonometric identities:
(a) cos 3θ = cos3 θ − 3 cos θ sin2 θ
(b) sin 3θ = 3 cos2 θ sin θ − sin3 θ

solution:
From de Moivre’s formula, we know: einθ = cos nθ + i sin nθ = (cos θ + i sin θ)n for n ∈ Z. Now, consider
the case of n = 3:

⇒ cos 3θ + i sin 3θ = (cos θ + i sin θ)3

= cos3 θ + 3 cos2 θ(i sin θ) + 3 cos θ(i sin θ)2 + (i sin θ)3
= (cos3 θ − 3 cos θ sin2 θ) + i(3 cos2 θ sin θ − sin3 θ)
⇒ cos 3θ = cos3 θ − 3 cos θ sin2 θ
sin 3θ = 3 cos2 θ sin θ − sin3 θ

Problem 8
√
Find (−8−8 3i)1/4 , express the roots in rectangular coordinates, exhibit them as the vertices of a certain
square, and point out which is the principal root.

solution:
√ 1
√
3
) = 16 exp i(− 23 π + 2kπ) where k ∈ Z
 
(−8 − 8 3i) =
√ 1/416(− 2 − i
2
= 2 exp i(− 16 π + k2 π) where k = 0, 1, 2, 3 ⇒ principal root is 2 exp i(− 16 π)
  
⇒ (−8 − 8 3i)

Now, we list all roots and plot them on rectangular coordinates:

√
√
 
1 3 1
⇒ c0 = 2 exp i(− π) = 2( − )= 3−i
6 2 2
√ √
 
1
c1 = c0 ∗ exp i π = ( 3 − i)i = (1 + i 3)
2
√ √
 
1
c2 = c1 ∗ exp i π = (1 + i 3)i = (− 3 + i)
2
√ √
 
1
c3 = c2 ∗ exp i π = (− 3 + i)i = (−1 − i 3) (see Figure 4)
2

Figure 7