If $z+\frac{1}{z}=2\cos\theta,$ where $z\in\Bbb C$, show that $\left|\frac{z^{2 n}-1}{z^{2n}+1}\right|=|\tan n\theta|$

Question

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 $n$?

Answer 1

$$z+\frac 1z=2\cos \theta\Rightarrow z^2-2\cos\theta z+1=0$$ Solve this quadratic to obtain $$z=cos\theta\pm i\sin\theta$$ Or, $$z=e^{i\theta},e^{-i\theta}$$ Now, $$\frac{z^{2n}-1}{z^{2n}+1}=\frac{z^n-1/z^n}{z^n+1/z^n}$$ In either of the above cases(for $z$), we have, $$z^n+1/z^n=e^{in\theta}+e^{-in\theta}=2\cos n\theta$$ But depending on the value of $z$ we have $$z^n-1/z^n=\pm \left(e^{in\theta}-e^{-in\theta}\right)=\pm 2\sin n\theta$$ We now have $$\left |\frac{z^{2n}-1}{z^{2n}+1}\right |=\left|\frac{\pm 2\sin n\theta}{2\cos n\theta}\right|=|\tan n\theta|$$ Hence proved.

Answer 2

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.

Answer 3

I think this method avoids solving quadratic equation

Take $|z|=r>0$ and let $z=re^{i\phi}=r(\cos \phi + i \sin \phi)$ for some $\phi \in [-\pi,\pi)$.

$$ z+\frac{1}{z} =re^{i\phi}+\frac{1}{re^{i\phi}}= re^{i\phi}+\frac{1}{r}e^{-i\phi}= r(\cos \phi + i \sin \phi)+\frac{1}{r}(\cos \phi - i \sin \phi)$$ $$=\left( r\cos \phi +\frac{1}{r} \cos \phi \right)+ i\left( r\sin \phi-\frac{1}{r}\sin \phi \right)=2\cos \theta $$ Then observe $Re\left(z+\frac{1}{z}\right)=2\cos \theta$ and $Im \left(z+\frac{1}{z}\right)=0 \implies r\sin \phi-\frac{1}{r}\sin \phi =0$

From this you can show that $$|z|=1\text{ and }(\cos \phi =\cos \theta \implies \phi\equiv \theta \text{ or } -\theta \mod 2\pi)\\ \therefore z=e^{i\phi}=e^{i\theta} \text{ or } e^{-i\theta} $$.

Then you have got the result