$f:\mathbb{C}\rightarrow\mathbb{C}$ is entire function such that $g(z)=f(1/z)$ has a pole at $z=0$, then is $f$ surjective?

Question

$f:\mathbb{C}\rightarrow\mathbb{C}$ is entire function such that $g(z)=f(1/z)$ has a pole at $z=0$, then is $f$ surjective? I can prove that $f$ will be a polynomial. and hence $f$ is surjective. am I right?

I suppose I should start posting answer as actual answers, even if they are very short. — mrf, Jan 15 '13 at 21:48
are you using the result that an entire function with pole of order $m$ at inifity must be apolynom of degree $m$? — Myshkin, May 03 '13 at 09:12

score 3 · Accepted Answer · answered Jan 15 '13 at 21:50

3

Yes, since your polynomial must have degree at least 1, this follows from the fundamental theorem of algebra.

(The fact that $g$ is a polynomial is proabably easiest to see from the Taylor series expansion of $f$ which immediately gives you a Laurent series expansion for $g$, but this is probably what you did already.)

answered Jan 15 '13 at 21:50

mrf

43,639

Sir, I can't see why being a degree $\ge1$ polynomial means it is surjective – Silent May 02 '19 at 13:33

$f:\mathbb{C}\rightarrow\mathbb{C}$ is entire function such that $g(z)=f(1/z)$ has a pole at $z=0$, then is $f$ surjective?

1 Answers1