WA 1: solutions
Problem 1. Find the real and imaginary parts of:
2
1−3i,(1 + i√3)6,1 + i
1−i5
, 1 + i√3
1−i!4
.
Solution. (a) 2
1−3i=2(1+3i)
(1−3i)(1+3i)=2(1+3i)
10 =1+3i
5=1
5+3
5i. Therefore,
Re 2
1−3i=1
5, Im 2
1−3i=3
5.
(b) (1+i√3)6=21
2+√3
2i6=2cos π
3+isin π
36= 26cos 6π
3+isin 6π
3=
64 cos(2π)+i·64 sin(2π) = 64 (we used de Moivre’s formula). Therefore, Re (1 + i√3)6=
64, Im (1 + i√3)6= 0.
(c) 1+i
1−i5=(1+i)2
(1−i)(1+i)5=2i
25=i5=i. Therefore, Re 1+i
1−i5= 0,
Im 1+i
1−i5= 1.
(d) 1 + i√3 = 2 1
2+√3
2i= 2 cos π
3+isin π
3= 2eiπ
3.
1−i=√21
√2−1
√2i=√2cos −π
4+isin −π
4=√2e−iπ
4. Then
1 + i√3
1−i!4
=2eiπ
3
√2e−iπ
44
=√2ei7π
12 4= (√2)4ei7π
3
= 4 cos 7π
3+isin 7π
3= 4 1
2+√3
2i!= 2 + 2√3i.
Therefore, Re 1+i√3
1−i4= 2, Im 1+i√3
1−i4= 2√3.
Problem 2. Find all complex zfor which (1 + z)(1 −z)−1is: (i) purely real;
(ii) purely imaginary.
Solution. We have w=1+z
1−z=(1+z)(1−z)
|1−z|2=1−|z|2+z−z
|1−z|2=1−|z|2+2iIm z
|1−z|2.
Clearly, wis undefined for z= 1. (i) wis purely real if and only if Im z= 0
and z6= 1, i.e., when z∈R\ {1}. (ii) wis purely imaginary if and only if |z|= 1
and z6= 1.
Problem 3. Prove the identity
|z1+z2|2+|z1−z2|2= 2(|z1|2+|z2|2)
and explain its geometrical meaning.
Solution. |z1+z2|2+|z1−z2|2= (z1+z2)(z1+z2)+(z1−z2)(z1−z2) =
z1z1+z2z1+z1z2+z2z2+z1z1−z2z1−z1z2+z2z2= 2z1z1+2z2z2= 2(|z1|2+|z2|2).
The identity means that, in the parallelogram built on the vectors representing
two complex numbers z1and z2, the sum of squares of the diagonals equal the sum
of squares of the sides (in the generic case, when the vectors are not parallel).
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