<s>[INST] Showing that if $P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0$, then x verify $(\frac{x+i}{x-i})^{2n+1} = 1$ I need to show that if $P_n(X)=\frac{(X+i)^{2n+1} -  (X-i)^{2n+1}}{2i} = 0$, then x verify $(\frac{x+i}{x-i})^{2n+1} = 1$.
Here is what i have done:
$$P_n(X)=\frac{(X+i)^{2n+1} -  (X-i)^{2n+1}}{2i} = 0 \\ (X+i)^{2n+1}-(X-i)^{2n+1} = 0*2i = 0 \\ (X+i)^{2n+1} = (X-i)^{2n+1} \\ \frac{(X+i)^{2n+1}}{(X-i)^{2n+1}} = 1 \\ \left(\frac{X+i}{X-i}\right)^{2n+1} = 1$$
Is my reasoning correct ?
 [/INST] Please see the discussion here where it is mentioned that $z=x+i$, $z^\ast=x-i$ and $z^{2n+1}={z^{\ast}}^{2n+1}$.  Another way to write this is $$\left(\frac{x+i}{x-i}\right)^{2n+1} = 1.$$
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