Answer to Question #282429 in Complex Analysis for iqa

Question

If z=cosθ+i sinθ,prove that

zn + z−n = 2cos(nθ), z^n − z^−n = 2i sin(nθ).

Answer 1

Use De Moivre's formula

$(\cos\theta+i\sin \theta)^n=\cos(n\theta)+i\sin(n\theta) , n\in \Z$

Then

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

$z^{-n}=(\cos\theta+i\sin \theta)^{-n}=\cos(-n\theta)+i\sin(-n\theta)$

$=\cos(n\theta)-i\sin(n\theta)$

$z^n+z^{-n}=\cos(n\theta)+i\sin(n\theta)$

$+\cos(n\theta)-i\sin(n\theta)$

$=2\cos(n\theta), n\in \Z$

$z^n-z^{-n}=\cos(n\theta)+i\sin(n\theta)-$

$-(\cos(n\theta)-i\sin(n\theta))$

$=2i\sin(n\theta), n\in \Z$