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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.

Donnie Darko
  • 501

1 Answers1

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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.

jjagmath
  • 18,214