Find all roots of the equation $(z-1)^{n}=(z+1)^{n}$.

What is the sum of the squares of the roots of this equation?

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Note that if $z=\cos 2 \varphi+i \sin 2 \varphi$, then $\frac{1+z}{1-z}=\frac{2 \cos ^{2} \varphi+2 i \sin \varphi \cos \varphi}{2 \sin ^{2} \varphi-2 i \sin \varphi \cos \varphi}=\frac{\cos \varphi}{-i \sin \varphi}=i \operatorname{ctg} \varphi$. Write the equation in the form $\left(\frac{z+1}{z-1}\right)^{n}=1$. Then $\frac{z+1}{z-1}=\sqrt[n]{1}=\cos \frac{2 \pi k}{n}+i \sin \frac{2 \pi k}{n}$, hence the roots of the equation

$z_{k}=i \operatorname{ctg} k \pi / n(1 \leq k \leq n-1)$.

On the other hand, expanding the brackets in the equation using the binomial formula and making simplifications, we bring it to the form $C_{n}^{n-1} z^{n-1}+C_{n}^{n-3} z^{n-3}+\ldots=0$. By Vieta's theorem, $\sigma_{1}\left(z_{1}, \ldots, z_{n-1}\right)=0$,

$\sigma_{2}\left(z_{1}, \ldots, z_{n-1}\right)=\frac{C_{n}^{n-3}}{C_{n}^{n-1}}=-\frac{(n-1)(n-2)}{6}$

From this, $z_{1}^{2}+z_{2}^{2}+\ldots+z_{n-1}^{2}=\sigma_{1}^{2}-2 \sigma_{2}=-\frac{(n-1)(n-2)}{3}$.

## Answer

$z_{k}=i \operatorname{ctg} \frac{k \pi}{n}(1 \leq k \leq n-1) ; \quad z_{1}^{2}+\ldots+z_{n-1}^{2}=-\frac{(n-1)(n-2)}{3}$

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