Exercise [09.05]

Sameed Zahoor · 23 May 2008, 11:43

The given transformation is
$z=\frac{-t+i}{t+i}$
It can also be written as,
$z=\frac{(-t+i)(t-i)}{(t+i)(t-i)}=\frac{1-t^2}{1+t^2}+i\frac{2t}{1+t^2}$ ...(1)
The unit circle in the z-plane is characterised by the equation
|z|=1

Hence z can be written as follows,
$z=\cos{\chi}+i\sin{\chi}=\frac{1-\tan^{2}{(\chi/2})}{1+\tan^{2}{(\chi/2)}}+i\frac{2\tan{(\chi/2)}}{1+\tan^{2}{(\chi/2)}}$ ...(2)
Comparing (1)&(2)we get,
$t=\tan{(\chi/2)}$

Exercise [09.05]

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