Let $F(z)$ be a meromorphic function on a domain $\Omega$ with a pole at $z=a .$ Let $C \subset \Omega$ be a simple closed contour with the point $z=a$ enclosed in its interior domain $I,$ and assume that $F(z)$ is continuous on $C .$ Prove that there exists a constant $M>0$ such that within $I, F(z)$ attains every value in the set $\{w \in \mathbb{C}:|w|>M\}$
My thought: How to use the equivalent condition $\lim_{z\rightarrow a}f(z)=\infty$ to that $a$ is a pole for $F(z)$.
Best Answer
Near $a$, you can write $f(z)$ as $\frac{g(z)}{(z-a)^n}$, for some $n\in\mathbb{N}$ ($n$ is the order of the pole) and some analytic function $g$ such that $g(a)\ne0$. Take $r>0$ such that $D(a,r)\setminus\{a\}\subset\Omega$ and that $z\in D(a,r)\setminus\{a\}\implies g(z)\ne0$. Then, in that punctered disk,$$\frac1{f(z)}=\frac{(z-a)^n}{g(z)}$$and this quotient defines an analytic function $\varphi\colon D(a,r)\longrightarrow\Bbb C$ if you put $\varphi=0$. By the open mapping theorem, $\varphi\bigl(D(a,r)\bigr)$ is an open set, and therefore it ontains some disk $D(0,\varepsilon)$. So, the image of $f$ contains every $z\in\Bbb C$ such that $|z|>\frac1\varepsilon$.