Solving an exponential equation, $ z^n = …$

Assuming $n$ is a whole, positive number. Calculate:

$$z^n = (z-1)^n.$$

Perhaps a hint on where to start will be appreciated.. I’m not looking for answers, but an explanation so that I can understand how to solve it!

Thanks

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Hint: Note that $z = 0$ cannot hold. Dividing by $z$ gives $\bigl(\frac{z-1}z\bigr)^n = 1$. So $\frac{z-1}z$ has to be an $n$-th root of unity, denote it by $\zeta$, then
$$\frac{z-1}z = \zeta \iff z-\zeta z = -1 \iff z = -\frac1{1-\zeta} $$

That’s not an exponential equation, but rewrite as
$$\left(\frac{z-1}{z}\right)^n = 1$$
and make a clever substitution. (Note that $z=0$ is not a solution, so it’s ok to divide.)

The answer is correct!

Here is a solution

$$ \left(\frac{z-1}{z}\right)^n = 1 = e^{i2\pi k} \implies \frac{z-1}{z}= e^{\frac{i2\pi k}{n}},\,k=1,\dots, n-1 $$

$$\implies z=\frac{1}{1-e^{\frac{i2\pi k}{n}}},\quad k=1,\dots,n-1. $$

Note: We did not consider the case $k=0$ because it corresponds to the case

$$z-1=z.$$