Pole

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

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

$<br /> g(z)=f(1/z)$
where $(z \not= 0)<br />$

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

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

any help would be appreciated, thank you :)

Quote:

Originally Posted by slevvio

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

$<br /> g(z)=f(1/z)$
where $(z \not= 0)<br />$

Show that g has a pole at 0.

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

great, thank you