<s>[INST] If $z+\frac{1}{z}=2\cos\theta,$ where $z\in\Bbb C$, show that $\left|\frac{z^{2 n}-1}{z^{2n}+1}\right|=|\tan n\theta|$ If $z+\frac{1}{z}=2 \cos \theta,$ where $z$ is a complex number, show that
$$
\left|\frac{z^{2 n}-1}{z^{2 n}+1}\right|=|\tan n \theta|
$$
My Approach:
$$
\begin{array}{l}|\sin \theta|=\left|\sqrt{1-\cos ^{2} \theta}\right| \\ =\left|\sqrt{1-\left(\frac{z^{2}+1}{2z}\right)^{2}}\right| \\ =\left|\sqrt{\frac{4 z^{2}-z^{4}-2 z^{2}-1}{4 z^{2}}} \right|\\ =\left|\sqrt{\frac{-\left(z^{4}-2 z^{2}+1\right)}{4 z^{2}}}\right|=\left|\sqrt{\frac{\left(z^{2}-1\right)^{2}}{4 z^{2}}}\right| \\ =|\frac{z^{2}-1}{2 z}|\end{array}
$$
So $|\tan \theta|=\left|\frac{z^{2}-1}{z^{2}+1}\right|$ ( proven when $n = 1$)
Is there any way to prove directly by taking $n$?
 [/INST] You could express the equation directly as a quadratic:
$z^2 - 2 z \cos \theta + 1 = 0$
and use the quadratic formula, for further insight.
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