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
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 $k=0,\dots, n-1$.
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,.
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.
