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I have some problems dealing with multivalued functions when it comes to handling singularities.

I'll give an example and try to ask questions based on it.

I want to classify the singularities of $f(z)=\sqrt z ^*$. I know it is defined as $f(z) =\exp ( \ln^* z)$. I use $*$ to indicate the complex version of the function. So $f(z) = \exp(\ln ( \sqrt{|z|} + i/2 \arg(z)+k \pi i)=\pm \exp(\ln ( \sqrt{|z|} + i/2 \arg(z))=:f_{\pm}(z)$.

$f_{\pm}$ are the two branches of $f$.

From here I don't know how to proceed. Is $0$ a singularity in the first place? I suppose it is, since one shouldn't be able to include $0$ in any convergence circle of a power (Taylor) series for $f_{\pm}$. Now, an idea would be to represent $f_-$ or $f_+$ in a Laurent series centered at 0: I could do that since $f_-$ and $f_+$ are continuous on a punctured disc centered in 0. Calculating the coefficients $a_n=\frac{1}{2 \pi i} \int_{|w|=R} f_+(w)/w^{n+1} dw=\frac{-2 \sqrt{R}}{(2n-1) \pi R^n}$ I conclude $0$ is essential for $f_+$, supposed I didn't commit stupid errors in evaluating the integral.

Furthermore why should it be correct to classify the singularities of $f$ by classifying those of its branches?

Anyobody can tell me what's wrong in my reasoning? Thanks in advance!

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  • $\begingroup$ There are no Laurent series for non meromorphic functions. With suitable definitions (analytic continuation) $\log(e^z)$ is analytic. A branch point at $z=0$ means $f(e^z)$ is analytic for $-Re(z)$ large enough. $\endgroup$ – reuns Oct 14 '18 at 18:21
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The classification of singularities is for isolated singularities. $z=0$ is not an isolated singularity of $\sqrt z$, since $\sqrt z$ is not holomorphic on any disc centered at $z=0$. Is is a branching point. The domain of the principal value branch of $\sqrt z$ is $\Bbb C\setminus(-\infty,0]$, and $0$ is not in that domain.

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