0
$\begingroup$

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 $z_0$.

$\endgroup$
  • $\begingroup$ is $U$ an open disk? The result seems untrue otherwise $\endgroup$ – user254433 Sep 9 '18 at 9:38
  • 1
    $\begingroup$ sorry, yes U is a open disk. $\endgroup$ – user270331 Sep 9 '18 at 9:41
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.