# Zeros of z^a-1

1. Jan 22, 2008

### birulami

What are the values of $z$ for which

$$z^\alpha-1=0$$

when $z$ and $\alpha$ are complex? Trivially,

$$r\cdot e^{k 2\pi i/\beta}$$

is a solution for every integer value $k$ if $r=1$ and $\beta=\alpha$.

Can every solution be written in this form? Again this is trivial for real $\alpha$, but how about $\alpha$ with a non-zero imaginary part? What are the values of $r$ and $\beta$ then?

Thanks,
Harald.

2. Jan 26, 2008

### sszabo

By definition
$$z^{a}=\exp (a\ln z)$$.​
Denote
$$u:=\ln z$$.​
So you have to solve
$$\exp(a u)=1$$.​
Hence
$$au=i2k\pi$$,​
where $$k$$ is arbitrary integer. It gives
$$u=\frac{i2k\pi}{a}$$.​
Now we solve
$$\ln z=\frac{i2k\pi}{a}=\frac{i2k\pi\bar{a}}{\left|a\right|^2}= \frac{2k\pi \text{Im}\,a}{\left|a\right|^2}+ \frac{i2k\pi\text{Re}\,a}{\left|a\right|^2}$$.​
If we find $$z$$ in the form $$z=r\exp(i\varphi)$$, then we obtain
$$\ln\,r=\frac{2k\pi \text{Im}\,a}{\left|a\right|^2}$$​
and
$$\varphi=\frac{2k\pi\text{Re}\,a}{\left|a\right|^2}+ 2\ell\pi$$​
where $$\ell$$ is an arbitrary integer.