Prove an equation in complex numbers Prove an equation ($n \in N$, $z \in C$, $a \in R$):
$$ z^n +  \frac1{z^n} = 2 \cos\alpha n$$
if $$ z +  \frac1{z} = 2 \cos\alpha$$
I tried math induction, but did not solve.

Using de Moivre's formula $(\cos \alpha + i \sin \alpha)^n = \cos n\alpha + i \sin n\alpha$, we see that setting $z = r(\cos \alpha + i\sin \alpha)$, we have $z^{-1} = r^{-1}(\cos \alpha - i \sin \alpha)$; thus
$z + z^{-1} = r(\cos \alpha + i \sin \alpha) + r^{-1}(\cos \alpha - i \sin \alpha) = 2 \cos \alpha. \tag{1}$
This implies
$ r\cos \alpha + r^{-1}\cos \alpha = 2 \cos \alpha, \tag{2}$
as well as
$i(r\sin \alpha -r^{-1}\sin \alpha) = 0, \tag{3}$
this latter equation (3) since $2 \cos \alpha$ is real.
If $\cos \alpha \ne 0$, (2) yields
$r + r^{-1} = 2, \tag{4}$
or 
$(r - 1)^2 = r^2 - 2r + 1 = 0; \tag{5}$
$r = 1$ and $z = \cos \alpha + \sin \alpha$; then $z^n = \cos (n\alpha) + i\sin (n\alpha)$,  $z^{-n} = \cos (-n\alpha) + i\sin (-n\alpha) = \cos (n\alpha) - i\sin (n\alpha)$, so
$z^n + z^{-n} = 2 \cos n \alpha \tag{6}$
in this case.
In the event that  $\cos \alpha = 0$, we have $\sin \alpha \ne 0$, and from (3) we infer
$r \sin \alpha = r^{-1} \sin \alpha, \tag{7}$
whence
$r^2 = 1, \tag{8}$
implying $r = 1$ since $r > 0$.  Thus we still have $z = \cos \alpha + i\sin \alpha$ and (6) still follows as in the case $\cos \alpha \ne 0$.  I credit Daniel Littlewood's comment for making clear to me the utility of equation (3) in this context.
Well, I hope that helps.  Cheers,
and as always,
Fiat Lux!!!
