Singularities of a function

Question

Given $f(z):=\operatorname{Log}(\frac{z-2}{z-3})$, $\operatorname{Log}:\mathbb{C}\setminus\mathbb{R_{\le 0}}\to \mathbb{C}$. Is in $z_0=3$ a essential singularity of f? I'm not sure what is correct... In my opinion $z_0$ is not isolated, there is no neighborhood $X=U\setminus \{3\}$ at 3 such that f is holomorphic on X. Therefore the residue of f at 3 has to be zero. But nevertheless, I calculated first: $$\operatorname{Log}(\frac{z-2}{z-3})=\operatorname{Log}(1+\frac{1}{z-3})=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{(z-3)^n*n}=\sum_{n=-\infty}^{-1}(-1)^{-n+1}\frac{(z-3)^n}{-n}$$ and I got $z_0=3$ is a essential singularity of f with $\operatorname{Res}(f,3)=-1$. Now I'm confused. Is it wrong to say that the coefficient of $z^{-1}$ is the residuum of f in 3 because in 3 f has no isolated singularity? Regards.

Answer 1

Is in $z_0 = 3$ a essential singularity of $f$?

That depends on what you call an essential singularity. If you call only isolated singularities that are neither removable nor poles essential singularities, then $3$ is not an essential singularity of $f$, since it's not an isolated singularity, you have the branch cut from $2$ to $3$. If you call every singularity that is neither removable nor a pole essential, isolated or not, then $3$ is an essential singularity of $f$.

Therefore the residue of $f$ at $3$ has to be zero.

No, it is not defined. The residue is only defined in isolated singularities.

Since $f$ is not holomorphic in a punctured neighbourhood of $3$, $f$ has no Laurent expansion in such a punctured neighbourhood. The Laurent expansion you computed is valid in the annulus $1 < \lvert z-3\rvert < \infty$, not in a punctured neighbourhood of $3$, and from such an expansion, you cannot read off a residue even if there was an isolated singularity in $3$. Consider for illustration $g(z) = \frac{1}{z(1-z)} = \frac{1}{z} + \frac{1}{1-z}$. $g$ has an isolated singularity in $0$, with residue $1$, but the Laurent expansion of $g$ in the annulus $1 < \lvert z\rvert < \infty$ is

$$\begin{align} g(z) &= \frac{1}{z} + \frac{1}{1-z}\\ &= \frac{1}{z} - \frac{1}{z}\cdot \frac{1}{1-\frac{1}{z}}\\ &= \frac{1}{z}\left(1-\sum_{k=0}^\infty \frac{1}{z^k}\right)\\ &= - \sum_{m=2}^\infty \frac{1}{z^m}, \end{align}$$

which has $0$ as the coefficient of $z^{-1}$.