1. ## Pole

Hello everyone I was wondering if anyone could give me a hand with this

Let f be an entire function such that $\displaystyle |f(z)| \rightarrow \infty$as |$\displaystyle z| \rightarrow \infty$. Define g by

$\displaystyle g(z)=f(1/z)$
where $\displaystyle (z \not= 0)$

Show that g has a pole at 0, and, by considering the Laurent series for g about 0, show that f is a polynomial function.

Im not so worried about the 2nd part yet but i was wondering how to show that g has a pole, intuitvely i can see that it has one at zero

i am trrying to express g as

$\displaystyle g(z) = \frac{h(z)}{z}$ where h(z) is analytic on some open disc around 0 and $\displaystyle k(0) \not= 0$
but im not sure how to do this

any help would be appreciated, thank you

2. Originally Posted by slevvio
Let f be an entire function such that $\displaystyle |f(z)| \rightarrow \infty$as |$\displaystyle z| \rightarrow \infty$. Define g by

$\displaystyle g(z)=f(1/z)$
where $\displaystyle (z \not= 0)$

Show that g has a pole at 0.
Let $\displaystyle h(z) = 1/g(z)\ (z\ne0)$. Then $\displaystyle h$ is analytic in some neighbourhood of 0 and tends to 0 as $\displaystyle z\to0$. So if we extend the domain of $\displaystyle h$ by defining $\displaystyle h(0)=0$ then $\displaystyle h$ is analytic at 0, and in fact has a zero of some order at 0. Therefore $\displaystyle g$ has a pole of that same order at 0.

3. great, thank you