How to formally show that $f(z)$ is analytic at $z=0$?

Question

Let $z$ be a complex number. Let $$f(z)=\dfrac{1}{\frac{1}{z}+\ln(\frac{1}{z})}.$$ How to formally show that $f(z)$ is analytic at $z=0$? I know that for small $z$ we have $$\left|\tfrac{1}{z}\right|>\left|\ln(\tfrac{1}{z})\right|$$ and that implies $|f(0)|=0.$ Are there multiple ways to handle this ?

Answer 1

To me, "analytic" means "locally represented by its Taylor series". With this interpretation $f$ is not analytic. Indeed, suppose $f(z)=z^r\sum_{n=0}^\infty c_n z^n $ in a neighborhood of $0$, where $c_0\ne 0$. Then $$\frac{1}{z}+\ln \frac{1}{z} = z^{-r} \sum_{n=0}^\infty b_n z^n $$ in some (possibly smaller) neighborhood of $0$. It follows that $\ln \frac{1}{z}$ has a pole or a removable singularity at $0$. If this is not evidently absurd already, apply the same to $\ln z=-\ln \frac{1}{z}$ and conclude that the logarithm is a rational function.

Answer 2

$$f(z)=\frac1{\frac1z+\text{Log}\frac1z}=\frac z{1+z\,\text{Log}\frac1z}$$

Now,

$$\text{Log}\frac1z:=\log\frac1{|z|}+i\arg\frac1z\implies z\,\text{Log}\frac1z=z\log\frac1{|z|}+iz\arg\frac1z$$

If you now choose a branch cut for the complex logarithm (and you better do if you have any hope to make any sense in this particular stuff), say the usual half negative real axis and zero ($\;\Bbb R_-\;$), you get a nice analytic function in the rest (and we don't care what happens with $\,z\to\infty\,$ since we're far away from there) , so

$$z\,\text{Log}\frac1z\xrightarrow[z\to 0\,,\,z\notin\Bbb R_-]{}0$$ and you can talk of a removable singularity of your function at $\,z=0\,$...although this would be a little absurd since you already removed this point by choosing the above branch cut for the logarithm, and this will be the situation with any branch cut you choose as any of them must contain the zero point...

In short, definitely not analytic there, though...