Prove equivalence of a complex function

Question

How to prove that,

$(z+1)^{100} = (z-1)^{100}$ is equivalent to $(z+1) = (z-1) e^\cfrac{2\times \pi \times k \times i}{100}$

Thank you.

Edit:

Okay sure, and I have attempted to do it.

First step is to take $(z+1)^{99}$ to the other side of the equation.

Then, $(z+1) = (z-1) \cfrac{(z-1)^{99}}{(z+1)^{99}}$

Then I believe the fraction should be equivalent to something??

Answer 1

Hint: Note that $$ \left(\frac{z+1}{z-1}\right)^{100}=1 $$ and that $e^{2\pi ki}=1$ for all $k\in\mathbb{Z}$.

Answer 2

Clearly $z\neq 1$, so $$ \begin{align*} (z + 1)^{100} = (z - 1)^{100} &\iff Z^{100} = 1 \quad\text{where}\quad Z = (z + 1)/(z - 1) \\ &\iff \begin{cases} \lvert Z^{100} \rvert = \lvert 1 \rvert \\ \arg(Z^{100}) \equiv \arg(1) \pmod{2\mathrm{\pi}} \\ \end{cases} \\ &\iff \begin{cases} \lvert {Z} \rvert^{100} = 1 \\ 100 \arg Z \equiv 0 \pmod{2\mathrm{\pi}} \end{cases} \\ &\iff \begin{cases} \lvert Z \rvert = 1 \\ \arg Z \equiv 0 \pmod{2\mathrm{\pi}/100} \end{cases} \\ &\iff Z = \mathrm{e}^{2\mathrm{i}k\mathrm{\pi}/100} \quad\text{where}\quad k\in\mathbb{Z}\\ &\iff z + 1 = (z - 1) \mathrm{e}^{2\mathrm{i}k\mathrm{\pi}/100} \quad\text{where}\quad k\in\mathbb{Z}. \end{align*} $$