Past exam of the mathematics course of the University of Cambridge / 2020 / ia / Paper 1 / 1C (Vectors and Matrices)

Table of contents
- i 1C
  - Solution i
- ii 1C
  - Solution ii
- iii 1C
  - Solution iii
- iv 1C
  - Solution iv
- v 1C
  - Solution v

i

Solution

Using Euler's formula, the complex exponential function satisfies

e^{x + i y} = e^{x} (cos y + i sin y) .

Hence

Re (e^{z}) = e^{x} cos y, Im (e^{z}) = e^{x} sin y .

ii

Solution

The complex sine function obeys

sin (x + i y) = sin x cosh y + i cos x sinh y .

Thus

Re (sin z) = sin x cosh y, Im (sin z) = cos x sinh y .

iii

Solution

Since

z \overline{z} = ∣ z ∣^{2} = x^{2} + y^{2}

, the complex conjugate gives

\frac{1}{z} - \frac{1}{z} = \frac{z - z}{∣ z ∣ ^{2}} = - \frac{2 i y}{x ^{2} + y ^{2}} .

Therefore

Re = 0, Im = - \frac{2 y}{x ^{2} + y ^{2}} .

iv

Solution

Factor the expression using the complex conjugate:

z^{3} - z^{2} \overline{z} - z \overline{z}^{2} + \overline{z}^{3} = (z + \overline{z}) (z - \overline{z})^{2} = (2 x) (2 i y)^{2} = - 8 x y^{2} .

It is real, so

Re = - 8 x y^{2}, Im = 0 .

v

Solution

All values of the complex logarithm are

w = lo g ∣ z ∣ + i (ar g z + 2 πn), n \in Z .

Because

x > 0

, one may take the complex argument

ar g z = arctan (y / x)

. Consequently

Re w = \frac{1}{2} lo g (x^{2} + y^{2}), Im w = arctan \frac{y}{x} + 2 πn, q u a d n \in Z .

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