Complex powers number of solutions For which complex $\alpha$ will $z^{\alpha}$ have precisely one value, more than one
but finitely many values, or infinitely many values? Justify your claims
I was thinking α=0 for one solutions cause then it would simply be 1
and $α=!0$ but alpha finite for finite solutions and then alpha infinite for infinitely many values?
I'm not sure if this is right though and would love to hear if anyone else has a solution?
 A: It is simplest to analyse when we write the solutions in the form $A^{\alpha}e^{i(x+2\pi n)\alpha}$ where $Ae^{i(x+2\pi n)} = z$ and $\alpha=a + bi$. where $n$ is an integer. 
Let's calculate $A^{\alpha}$, which just becomes $A^aA^{bi}$. The second part becomes $e^{-b(x+2\pi n)}e^{ia(x+2\pi n)}$
So $A^aA^{bi}e^{-b(x+2\pi n)}e^{ia(x+2\pi n)}$ is the result. Gather the magnitudes and complex parts to give $A^ae^{-bx}e^{-2\pi b n} = |z^\alpha|.$ Also, $arg(z^\alpha) = a(x + 2\pi n)+blog(A)$ for any integer $n$. 
Note that there are infinite solutions in magnitude for the result using any $z$ if $b$ is non zero. Also, it's interesting that there are an infinite number of solutions (distinct) in argument for $z^\alpha$ if and only if $a$ is irrational, because $a\cdot{n}$ never repeats for any $n$. I find it particularly strange that every solution in magnitude is governed by the same variable ($n$) as the argument. So when $a$ is irrational, each argument also has a unique solution in magnitude, whereas when $a$ is rational, each argument solution has an infinite number of corresponding magnitudes (that is, given $b$ is non zero). Of course, when $a$ is an integer we only have one argument.
I hope my notation was clear enough, take care to not confuse $\alpha$ with $a$. 
