For what values $\alpha$ for complex z $\ln(z^{\alpha}) = \alpha \ln(z)$?

For example, when $\alpha = 2$, $\ln(z^{2}) \neq 2\ln(z)$, because argument z is determined up to constant $2 \pi k$. So

$$ \ln(z^{2}) = \ln(z) + \ln(z) = \ln(z_{k_{1}}) + \ln(z_{k_{2}}) \neq 2\ln(z_{k_{3}}). $$

Of course, two of correct answers - $1, -1$. But I know, that there are an infinite number of answers for $\alpha$. Can you help me?

Using this and this,

$Log (a+ib)^{x+iy}$ $=\frac{1}{2}x\log (a^2+b^2)-y(2m\pi+\tan ^{-1}\frac{b}{a})+i(\frac{1}{2}y\log (a^2+b^2)+x(2m\pi+\tan ^{-1}\frac{b}{a}))$

$Log (a+ib)= \frac{1}{2}\log (a^2+b^2)+i(2n\pi+\tan ^{-1}\frac{b}{a})$

$(x+iy)Log (a+ib)$ $=\frac{1}{2}x\log (a^2+b^2)-y(2n\pi+\tan ^{-1}\frac{b}{a})+i(x(2n\pi+\tan ^{-1}\frac{b}{a})+\frac{1}{2}y\log (a^2+b^2))$

The principal values will be same if $-\pi<\frac{1}{2}y\log (a^2+b^2)+x\tan ^{-1}\frac{b}{a}\le \pi$ else $2r\pi$ (where $r$ is any integer) must be added to this argument to adjust its value in $(-\pi, \pi]$.

If $x=0$ or $(b=0$ and $a>0 ⇔ \tan ^{-1}\frac{b}{a}=0)$ , the condition becomes $-\pi<\frac{1}{2}y\log (a^2+b^2)\le \pi$

If $y=0$ or $a^2+b^2=1 ⇔ \log (a^2+b^2)=0$, the condition becomes $-\pi<x\tan ^{-1}\frac{b}{a}\le \pi$

If $y=0$ or $a^2+b^2=1$ and $x=-1$ the condition becomes $-\pi<-\tan ^{-1}\frac{b}{a}\le \pi\implies \pi>\tan ^{-1}\frac{b}{a}\ge -\tan ^{-1}\frac{b}{a}$ which is true for all $a,b$.

If $y=0,x=2$ the condition becomes $-\pi<2\tan ^{-1}\frac{b}{a}\le \pi $

$\implies -\frac{\pi}2<\tan ^{-1}\frac{b}{a}\le \frac{\pi}2$ i.e, $a>0$.