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

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

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

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

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

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

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

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

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

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