I'm working on a problem from Complex Variables by Taylor, in the section about Liouville's Theorem: Show that if $f$ is an entire function and $\lim_{z \rightarrow \infty} f(z) = \infty$, then $f$ must have a zero somewhere in $\mathbb{C}$.
I can see this amounts to showing that $f(z)$ is a polynomial and then use the Fundamental Theorem of Algebra, and the problem also states to use the previous exercise, the result of which is that if $\lim_{z \rightarrow \infty} f(z) = \infty$, then $\lim_{z \rightarrow \infty} \frac{1}{f(z)} = 0$. I can't really see how to get started except for maybe taking the power series expansion of $f$?
Any advice?
Best Answer
By contradiction, assume that $f$ does not have a zero in $\mathbb{C}$. Since $\lim_{z\to\infty}f(z)=\infty$, $$|f(z)|\geq C_1\mbox{ in }\mathbb{C}-B_R(0)$$ for some constant $C_1>0$. On the other hand, since $f$ does not have a zero in $\mathbb{C}$, $f$ does not have a zero in $\overline{B_R(0)}$. Since $\overline{B_R(0)}$ is compact, we have $$|f(z)|\geq C_2\mbox{ in }\overline{B_R(0)}$$ for some constant $C_2>0$. Combining this, we have $\frac{1}{f}$ is an entire function such that $$\left|\frac{1}{f}\right|\leq\frac{1}{\min\{C_1,C_2\}}$$ which is bounded. By Liouville's Theore, $\frac{1}{f}$ is constant, which contradicts to the fact that $\lim_{z\to\infty}f(z)=\infty$.