MA2325: SOLUTIONS TO ASSIGNMENT 1
1. Compute the modulus for the following complex numbers.
(a)
2+3i
1−4i
(b)
1+i
√
2
7
Solution:
q
13
(a)
17
7
1+i
√
√
(b) | 1+i
|
=
|
|7 = 1
2
2
2. Solve the following equations.
(a) z 3 = 1
(b) z 4 = 16i
Solution:
An equation of the form z n = w with w 6= 0 has n
distinct solutions. Let z have trigonometric form z = |z|(cos φ +
i sin φ). Then z n = |z|n (cos nφ + i sin nφ) (de Moivre). Now |z|n =
1
|z n | = |w| and so |z| = |w| n . We have nφ ∈ {Arg(w) + 2πj : j ∈
Z} where Arg(w) is the principle value of the argument of w. In
particular φ ∈ { Arg(w)+2πj
: j ∈ Z}. Let φj =
n
Arg(w)+2πj
n
for each
j ∈ Z. Note that if φj and φk differ by an integer multiple of 2π
then they produce the same solution. The n distinct solutions are
1
zj = |w| n (cos φj + i sin φj )
1
Arg(w) + 2πj
Arg(w) + 2πj
= |w| n cos
+ i sin
n
n
where j = 0, 1, . . . , n − 1.
(a) The solutions are z0 = 1, z1 = − 21 +
1
√
3
i
2
and z2 = − 12 −
√
3
i.
2
2
MA2325: SOLUTIONS TO ASSIGNMENT 1
1.0
0.5
-1.0
0.5
-0.5
1.0
-0.5
-1.0
Figure 1. Solutions for z 3 = 1
2
1
-2
1
-1
2
-1
-2
Figure 2. Solutions for z 4 = 16i
(b) The solutions are
π
z0 = 2 cos
8
5π
z1 = 2 cos
8
9π
z2 = 2 cos
8
13π
z3 = 2 cos
8
π
8
5π
+ i sin
8
9π
+ i sin
8
13π
+ i sin
8
+ i sin
3. State whether the following sets are open, closed, bounded or connected.
(a) {z ∈ C : |z − (1 + 2i)| < 4}
(b) {z ∈ C : |z − (3 + 2i)| = 1} ∪ {z ∈ C : |z − (2 − i)| = 1}
MA2325: SOLUTIONS TO ASSIGNMENT 1
3
(c) {z ∈ C : 0 < Re(z) < 1 and Im(z) = 0}
(d) {1 + i, −1 − i, −1 + i, 1 − i}
Solution:
(a) Open, not closed, bounded, connected.
(b) Not open, closed, bounded, not connected.
(c) Not open, not closed, bounded, connected.
(d) Not open, closed, bounded, not connected.
4. Prove that the following functions are continuous on C.
(a) z 7→ Re(z)
(b) z 7→ Im(z)
(c) z 7→ z̄
(d) z 7→ |z|
Solution:
To prove a function f is continuous we need to show
it is continuous at every point in its domain. In this question the
domain of each function is C. Let w ∈ C. We need to show that
given any positive real number there exists a positive real number
δ such that the following implication holds,
0 < |z − w| < δ =⇒ |f (z) − f (w)| < (a) Given any positive real number we can choose δ = since
0 < |z − w| < =⇒ |Re(z) − Re(w)| = |Re(z − w)| ≤ |z − w| < (b) Given any positive real number we can again choose δ = since
0 < |z − w| < =⇒ |Im(z) − Im(w)| = |Im(z − w)| ≤ |z − w| < 4
MA2325: SOLUTIONS TO ASSIGNMENT 1
(c) Given any positive real number we can choose δ = since
0 < |z − w| < =⇒ |z̄ − w̄| = |z − w| = |z − w| < (d) Given any positive real number we can choose δ = since
0 < |z − w| < =⇒ ||z| − |w|| ≤ |z − w| < Here we used the reverse triangle inequality.