Complex root argument justification Show that the roots of ${(1+z)^n}$ = ${(1 -z)^n}$ are the values of $$i \tan (rπ/n)$$ where $$r = 0, 1, 2, ... n - 1,$$ but omitting n/2 if n is even.
I have got the expression of root I just want a concrete explanation of why $\frac{n}{2}$ can be omitted when n is even. The thing i deduced was that when n is even the expression has $(n-1)$ roots with 0 being one so the rest $(n-2)$ roots must be complex conjugate pairs,so exclude n/2. But I don't think it's concrete or the right reason. Also if i include $\frac{n}{2}$ in $n$ being odd case
then there's no complex conjugate of it $i(\infty)$ so how the integer coefficient is restored. Please explain the case of $\frac{n}{2}$ for both the cases. I don't want any worked out process of answer but just to enlighten me on my confusion and doubt . Please help.
 A: Even case
You need to omit $n/2$ if $n$ is even otherwise the tangent diverges (at $ \pi/2$) indeed you will have :
$$r=n/2$$
so
$$ x_r=i\tan(r\pi/n)=i\tan(\pi/2)=\pm\infty$$
and you work to find finite complex number.
When you use a reasonement using analysis-synthesis

*

*You first suppose that you have solution, you find a certain form here the form you write

*You have to verify that they all work or otherwise exclude the case that don't work (for example here $r=n/2$ when $n$ is even)

Odd case
You want to avoid infinite values so if $n$ is odd, we have to look at :
$$ i\tan(r\pi/n) $$ when $r$ goes from $0$ to $n-1$
Suppose that there exists $r \in [|0,n-1|], \ r\pi/n=(2k+1)\pi/2 , \ k \in \mathbb{Z}$ (it is the case of divergence i.e. $\tan(r\pi/n)=\infty$)
You would have $r=n(2k+1)/2$ which is not whole number because $n$ and $(2k+1)$ are odd so their product divided by 2 is not whole.
So it is not possible to have $\tan(r\pi/n)=\infty$ is the case of $n$ odd.
Hope it helps you to see clearer.
