# Classification of singularities of complex multivalued function

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!

• 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. – reuns Oct 14 '18 at 18:21

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.