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.