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I have some trouble with the concept of removable singularities in complex analysis:

I understand, that if $\lim_{z\rightarrow x} (z-x) f(z) = 0$ holds, then $f$ has a removable singularity and can be continously extended to $z=x$, i.e. $f(x)$ takes a finite value.

Now, take the function $f_a(z) = z^{-a}$ for $a \in (0,1)$, which is undefined for $z=0$, but holomorphic on the whole $\mathbb{C} \setminus \{0\}$, right? How does it holomorphic extension look like? What is the value that can be assigned to $f_a(0)$?

Thanks for your help!

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    $\begingroup$ In that particular example you have a bigger problem, which is that the exponential is only defined with a branch cut. This branch cut makes it impossible to remove the singularity at $0$. $\endgroup$
    – Ian
    Jun 13, 2015 at 0:20
  • $\begingroup$ Why do yo think $z=0$ is a removable singularly of $z^{-\alpha}$? $\endgroup$ Jun 13, 2015 at 0:21

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It turns out $z^{-1/2}$ is not holomorphic in $\mathbb{C}\setminus \{0\}$. In little neighborhoods which don't contain $0$, you can pick one of the two branches of the square root function. Then it's ok. So we're gonna have problems making non-integer power functions into entire functions.

It's easier to show that $z^{1/2}$ isn't holomorphic in any neighborhood that contains a loop around $0$. The idea is to take a loop around 0 and follow it all the way around. Consider

$$ \gamma(t) = \alpha e^{it} $$ for small enough $\alpha >0$ and $t \in [0,2\pi]$. Then $\sqrt{\gamma(0)} = \sqrt{\alpha}$. Note we could have chosen $-\alpha$ here. This will be a problem in a moment. If we assume $\sqrt{\gamma(t)}$ is continuous, it'll have to be $$ \sqrt{\gamma(t)}=\sqrt{\alpha}e^{it/2} $$

So $\sqrt{\gamma(2\pi)} = -\alpha$. This is bad, because $\gamma(2\pi)=\gamma(0)$. but their square roots are different.

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