Singular point of $f(z)$ also a singular point of $1/f(z)$ and $f^{2}(z)$ Suppose $z_{0} \in \mathbb{C}$ is an isolated singular point of the function $f$ of a given type (removable, pole of order $N$, essential). I need to show that $z_{0}$ is an isolated singular point of 


*

*$g(z) = 1/f(z)$ (here, additionally, assume that $f(z)$ has no zeros in some deleted neighborhood of $z_{0}$).

*$h(z) = f^{2}(z)$


and find its type in each case.
I've been struggling with this problem for a couple of days - I tried using the neighborhood definition of an isolated singular point with $\delta$'s, but that wasn't getting me anywhere. 
Could somebody please walk me through this problem? I'm extremely confused to the point of tearing my hair out. Maybe just a full solution for 1. and a hint for 2.? 
Thanks.
 A: If $f$ has no zeroes in some neighbourhood $U$ of $z$, then $f(U-\{z\})$ is some subset of $\mathbb C$ that does not contain $0$. The inversion $\iota: x \mapsto \frac1x$ is continuous (holomorphic if the set is open) on such a subset. This means that $\iota \circ f: U - \{z\} \to \mathbb C$ is holomorphic as a composition of holomorphic functions. But $(\iota \circ f)(x)$ is just $\frac1{f(x)}$. ($f(U-\{z\})$ is an open set since holomorphic maps map open sets to open sets).
So $\frac1{f(x)}$ is holomorphic on $U-\{z\}$ for some neighbourhood of $z$, this means the singularity is isolated.
In order to see what kind of singularity comes out start with the case of a pole.


*

*Pole -  This means that you have a neighbourhood $U$ of $z$ so that on $U-\{z\}$ the function $f$ is of the form $f(x)=(x-z)^{-n} g(x)$ for some holomorphic function $g$ defined on all of $U$ and that is not zero in $U$. The inversion is $(\iota \circ f)(x)=(x-z)^n \frac1{g(x)}$ and is a priori defined only on $U-\{z\}$.
But since $\frac1g$ is holomorphic and defined on all of $U$ (it has no zeros), and the polynomial $(x-z)^n$ also is holomorphic on all of $U$, the function $\frac1{f(x)}$ can be extended holomorphically by $\frac{(x-z)^n}{g(x)}$ onto all of $U$.

*Removable - There are two subcases, either the value of the extension of $f$ at $z$ is $0$ or it is not $0$. If it is not $0$ then $f$ is the restriction of a holomorphic function $\tilde f: U \to \mathbb C$ that does not take the value $0$ on $U-\{z\}$. But then $\frac1{\tilde f}$ is holomorphic on $U$ and $\frac1f=\frac1{\tilde f}\large\lvert_{U-\{z\}}$ is extended by $\frac1{\tilde f}$ and the singularity is removable.
If $\tilde f$ is the extension of $f$ onto $U$ and $\tilde f(z)=0$, then it is true that $\tilde f(x)=(x-z)^n g(x)$ where $g$ is holomorphic and does not have any zeros on $U$. This can be seen by considering the power series expansion of $\tilde f$ at $z$: $\tilde f(x)=\sum_k a_k (x-z)^k$. If $\tilde f(z)=0$ then $a_0=0$ and maybe some $a_k=0$ for $k\in\{0,..,n\}$ for some $n$. Thats why you can write $\tilde f(x)=(x-z)^n \sum_{k=n} a_k (x-z)^{k-n}$ with $a_n\neq 0$. At any rate this shows that $\tilde f$ can be put in the form $f(x)=(x-z)^n\ g(x)$ where $g(z)\neq 0$ and $g$ is holomorphic (it has an absolutely convergent power series) in some neighbourhood of $z$. By shrinking the domain even further you can assume $g$ is not zero anywhere in the domain. This means $\frac1{g}$ is holomorphic.
Putting this part together $f(x)=\tilde f(x)\lvert_{U-\{z\}} = (x-z)^n g(x)$ with $g$ nonzero and holomorphic. So $\frac{1}{f(x)}=(x-z)^{-n}\frac{1}{g(x)}$ with $\frac1{g}$ being holomorphic at $z$ and non-zero. This means $\frac{1}f$ has a pole at $z$.

*Essential - A singularity is essential if it is neither a pole or a removable singularity. Now we know that $\iota$ takes functions with poles at $z$ and makes them removable singularities. On the other hand it takes removable singularities and and makes them either poles or removable. Note now that $(\iota \circ \iota \circ f)=f$ on $U-\{z\}$. So if $\frac1f$ has a removable singularity or a pole at $z$, then $f=\frac1{1/f}$ has a removable singularity or a pole at $z$, ie it does not have an essential singularity.
However since $\frac1f$ definitely has a singularity at $z$, if $f$ is essential so $\frac1f$ must also be essential.
For case (2) the same argument goes through, but you must use a function other than $\iota$.
