$(z-1)^n+ (z+ 1)^n=0$? Let $n$ be a positive integer, and let $z∈\mathbb{C}$ satisfy $(z-1)^n+ (z+ 1)^n=0$.
a, I have to show that $z = (1+w)/(1-w)$, where $w^n = -1$
b, Show that $w \bar w=1$
c, Deduce that $z$ lies on the imaginary axis.
I got the a part by rearranging for $z$, but I'm stuck for part b and c.
I know that $w \bar w=1$ means that $w=\cos(a) + i \sin(a)$, but I don't know how to use it.
Thanks for any helps or hints 
 A: Suppose we have 
$$\begin{align}
\left(\frac{z-1}{z+1}\right)^n&=-1\\\\
&=e^{i(2k+1)\pi}
\end{align}$$
for any integer $k$.  
Then, we have 

$$\frac{z-1}{z+1}=e^{i(2k+1)\pi/n} \tag 1$$

for $k=0,\dots, n-1$.  
Solving $(1)$ for $z$ reveals
$$\begin{align}
z&=\frac{1+e^{i(2k+1)\pi/n}}{1-e^{i(2k+1)\pi/n}}\\\\
&=\bbox[5px,border:2px solid #C0A000]{-i \cot\left(\frac{(2k+1)\pi}{2n}\right)}
\end{align}$$
for $k=0,\dots, n-1$.
A: The part (c) can be deduced without calculating the solutions:
$$(z−1)^n + (z+1)^n = 0,$$
$$(z−1)^n = -(z+1)^n,$$
$$|z−1|^n = |z+1|^n,$$
$$|z−1| = |z+1|.$$
I.e., the solutions are equidistant of $1$ and $-1$, that is to say the are on the imaginary axis. Warning: not all the points of the imaginary axis are solutions,.
A: You can do it with elementary arithmetic of complex numbers. You only need the multiplicativity of the complex absolute value and basic knowledge about complex conjugation, like $w\overline w = \lvert w \rvert^2$.
a) goes by algebra, as you already did.
b) From $w^n = -1$ you have $\lvert w \rvert^n = 1$, so $w\overline w = \lvert w \rvert^2 = 1^2 = 1$ (since $1$ is the only nonnegative real number $x$ with $x^n = 1$).
c) From $1 = w\overline w$, using a) and c),
$$\overline z = \frac{1 + \overline w}{1 - \overline w} = \frac{w\overline w + \overline w}{w\overline w - \overline w} = \frac{w + 1}{w - 1} = -z,$$
so $z$ must be imaginary.
