complex numbers – NST First Year Maths

2016 Paper 1 Question 11

8th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(a)

When $\lambda = 1$

$|z-i| = |z^*-i|$

Geometrically this means $z$ is as far from $i$ as $z^*$ .

Let $z = x +i\,y$

$\begin{aligned} |x+iy-i| &= |x-iy-i| \\ \Rightarrow x^2 + (y-1)^2 &= x^2 + (y+1)^2 \\ \Rightarrow y^2-2y+1 &= y^2 +2y +1 \\ \Rightarrow y &= 0 \end{aligned}$

$y=0$ is the equation of the straight line in the complex plane. This is consistent with the geometric interpretation.

(b)

When $\lambda = 0$ it must be that $z=i$ . This is another straight, horizontal, line.

2014 Paper 1 Question 12

6th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(b)

$\tanh z = -i$

$\begin{aligned} \Rightarrow \frac{\sinh z}{\cosh z} &= -i \\ \Rightarrow \frac{e^z-e^{-z}}{e^z+e^{-z}} &= -i \\ \Rightarrow e^{2z}-1 &= -i \, ( e^{2z}+1 ) \\ \Rightarrow (1+i)e^{2z} &= 1 - i \\ \Rightarrow e^{2z} &= \frac{1-i}{1+i} \\ \Rightarrow e^{2z} &= \frac{1}{2}(1-2i-1) \\ \Rightarrow e^{2z} &= -i \\ \Rightarrow e^{2z} &= e^{-i\left( \frac{\pi}{2} +2k\pi \right) }, \quad k \in \mathbb{Z} \\ \Rightarrow 2z &= -i\left( \frac{\pi}{2} +2k\pi \right) \\ \Rightarrow z &= -i\left( \frac{\pi}{4} +k\pi \right), \quad k \in \mathbb{Z} \\ \end{aligned}$