Complex Solutions to z^\alpha-1=0

[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.

 

[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

$$\ln z=\frac{i2k\pi}{a}=\frac{i2k\pi\bar{a}}{\left|a\right|^2}=<br /> \frac{2k\pi \text{Im}\,a}{\left|a\right|^2}+<br /> \frac{i2k\pi\text{Re}\,a}{\left|a\right|^2}$$.​

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

$$\varphi=\frac{2k\pi\text{Re}\,a}{\left|a\right|^2}+<br /> 2\ell\pi$$​

where $$\ell$$ is an arbitrary integer.