# If $p\circ f$ has a pole at $z_0$ then so does $f$, where $p$ is a polynomial

Let $f : U$ \ ${{z_0}} \to C$ be a holomorphic function, where $U$ is an open disk and let $p$ be a non-constant polynomial. Show that the singularity of $f$ at $z_0$ is a pole of $f$ iff it is a pole of $p \circ f$.

I have proved the 'only if' part of the above question but i am having trouble in proving the 'if' part. Suppose $p\circ f$ has a pole at $z_0$ i.e. There is a holomorphic function $g: U\to C$ with $g(z_0) \neq 0$ such that $$p\circ f = \frac{g(z)}{(z-z_0)^n}$$

So, how should I conclude that f has a pole at $z_0$.

• is $U$ an open disk? The result seems untrue otherwise – user254433 Sep 9 '18 at 9:38
• sorry, yes U is a open disk. – user270331 Sep 9 '18 at 9:41

## 1 Answer

Hint: $f$ has a pole at $z_0$ if and only if $lim_{z\to z_0}|f(z)|=\infty$ and $f$ has an essential season singularity at $z_0$ if and only if the image of $B(z_0,\delta)\setminus \{z_0\}$ under $f$ is dense in $\mathbb{C}$ for all $\delta$ small enough.