So I've got this question that is a bit difficult to ask, since it uses a term in my language that I can't properly translate into English.

For $z\in\mathbb{C}^*$ and $a\in\mathbb{C}$ it would be natural to define $z^a=exp(a\cdot log(z))$, with $log(z)$ being a complex number so that $exp(log(z))=z$. I have a problem however, since $log(z)$ isn't single-valued. I'm asked to give combinations of $a$ and $z$ for which the definition is single-valued.

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    $\begingroup$ wouldn't this just involve choosing a particular branch for the $\log(z)$? $\endgroup$ – gt6989b Mar 10 '14 at 12:14
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    $\begingroup$ $\log$ is injective (with the usual mathematical English meaning of the word), what it isn't is single-valued. You can make $\log$ single-valued by, for example, not defining it for non-positive real numbers. $\endgroup$ – Gerry Myerson Mar 10 '14 at 12:19
  • $\begingroup$ I believe 'single-valued' is the term I was looking for, thanks! So, does that mean that $z^a$ is single-valued for any $a$, as long as $z \in \mathbb{R}\subset\mathbb{C}$? $\endgroup$ – RBS Mar 10 '14 at 12:42
  • $\begingroup$ No, for example $x^{1/2}$ has multiple solutions, eg $(-1)^{1/2} = \pm i$. It's even worse for irrational exponents. $\endgroup$ – Najib Idrissi Mar 10 '14 at 12:49
  • $\begingroup$ Ok, so what would be a sufficient condition for $a$ and $z$ for which $z^a$ is single-valued? I'm a bit at a loss here.. $\endgroup$ – RBS Mar 10 '14 at 13:10

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