If $f(z) = \sum_{n=1}^\infty a_nz^n$ is an entire function which does not have an essential singularity at $\infty$, then $f$ is a polynomial.
I'm trying to apply Liouville's Theorem, but I am not guaranteed that $f(z)$ is bounded.
Asked
Active
Viewed 64 times
1
-
1What does it mean for $f(z)$ to not have an essential singularity at $\infty$? Write out this definition EXPLICITLY. – peek-a-boo Apr 22 '21 at 21:38
-
Once you know that $f(z)=O(1+|z|^k)$ then $g(z)=\frac{f(z)-\sum_{n=1}^k a_n z^n}{z^{k+1}}$ is an entire function vanishing at $\infty$ so $|g(z)|$ attains its maximum at some $c\in \Bbb{C}$ which implies that all the $g^{(n)}(c),n\ge 1$ vanish ie. $g$ is constant, $=0$. – reuns Apr 22 '21 at 22:10