Question

About the complex number​

Answer 1

We're given

`z = \mathrm{cis}(\theta) = \cos(\theta) + i \sin(\theta)`

By de Moivre's theorem,

`z^n = \cos(n\theta) + i \sin(n\theta)`

so that

`(z^2 - 1)/(z^2 + 1) = (\cos(2\theta) + i \sin(2\theta) - 1)/(\cos(2\theta) + i \sin(2\theta) + 1)`

Multiply uniformly by the conjugate of the denominator.

`(z^2 - 1)/(z^2 + 1) = (\cos(2\theta) + i \sin(2\theta) - 1)/(\cos(2\theta) + i \sin(2\theta) + 1)\cdot(\cos(2\theta) - i \sin(2\theta) + 1)/(\cos(2\theta) - i \sin(2\theta) + 1)`

`(z^2 - 1)/(z^2 + 1) = (\cos^2(2\theta) - 1 + 2i\sin(2\theta) + \sin^2(2\theta))/((\cos(2\theta)+1)^2 + \sin^2(2\theta))`

`(z^2 - 1)/(z^2 + 1) = (2i\sin(2\theta))/(\cos^2(2\theta) + 2\cos(2\theta) + 1+ \sin^2(2\theta))`

`(z^2 - 1)/(z^2 + 1) = (2i\sin(2\theta))/(2\cos(2\theta) + 2)`

`(z^2 - 1)/(z^2 + 1) = (i\sin(2\theta))/(\cos(2\theta) + 1)`

`(z^2 - 1)/(z^2 + 1) = (2i\sin(\theta)\cos(\theta))/(2\cos^2(\theta) - 1 + 1)`

`(z^2 - 1)/(z^2 + 1) = (i\sin(\theta))/(\cos(\theta))`

`(z^2 - 1)/(z^2 + 1) = i\tan(\theta)`

QED