solve the equation $\left(z^{2}+1\right)^{n}-(z-1)^{2 n}=0$ Im trying to solve the eq $$(E)\quad \left(z^{2}+1\right)^{n}-(z-1)^{2 n}=0$$
My attemp :
By Newton i get
$$\left(z^{2}+1\right)^{n}-(z-1)^{2n}=0 \Leftrightarrow \sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) z^{2 k}-\sum_{k=0}^{2n}\left(\begin{array}{l}
n \\
k
\end{array}\right) z^{k}=0$$
which leaves us with : $$-\left(\begin{array}{l}
n \\
1
\end{array}\right) z-\left(\begin{array}{c}
n \\
3
\end{array}\right) z^{3}-\left(\begin{array}{c}
n \\
5
\end{array}\right) z^{5} \cdots-\left(\begin{array}{c}
2n \\
2n-1
\end{array}\right) z^{n}=0$$.I got stuck there.

Since $(z^2+1)^n -(z-1)^{2n}=0$ we have $(z^2+1)^n = ((z-1)^2)^n$, then $$\left(\frac{z^2+1}{(z-1)^2}\right)^n = 1$$
That means $\frac{z^2+1}{(z-1)^2} = \xi$, an $n$-th root of unity. Now just solve the quadratic. You'll get $2n-1$ solutions, $2$ for each $n$-th root of unity with the exception of $\xi = 1$ which gives only one solution.
