UMD Analysis Qualifying Exam/Jan09 Complex

Define new function h(z)
Define $$ h(z)= \frac{f(z)}{g(z)} \!\,$$.

h is continuous on the closure of D
Since $$  g \neq 0  \!\,$$ on $$ D   \!\,$$, then by the Maximum Modulus Principle, $$      g \!\,$$ is not zero in $$  \overline{D}           \!\,$$.

Hence, since $$ f              \!\,$$    and  $$        g        \!\,$$ are analytic on $$    \overline{D}            \!\,$$   and $$  g \neq 0     \!\,$$ on $$   \overline{D}             \!\,$$,     then     $$     h  \!\,$$ is analytic on $$   \overline{D}             \!\,$$ which implies $$       h    \!\,$$ is continuous on $$  \overline{D}        \!\,$$

h is analytic on D
This follows from above

Case 1: h(z) non-constant on D
If $$    h        \!\,$$ is not constant on $$   \overline{D}       \!\,$$, then by Maximum Modulus Principle, $$                |h| \!\,$$ achieves its maximum value on the boundary of $$    D  \!\,$$.

But since $$ |h(z)| \leq 1       \!\,$$ on $$   \partial D  \!\,$$ (by the hypothesis), then

$$ |h(z)| \leq 1     \!\,$$ on $$   \overline{D}             \!\,$$.

In particular $$ |h(0)| \leq 1              \!\,$$, or equivalently

$$  |f(0)| \leq |g(0)|             \!\,$$

Case 2: h(z) constant on D
Suppose that $$h(z) \!\,$$ is constant. Then

$$|h(z)|=\left|\frac{f(z)}{g(z)}\right| = |\alpha| \!\,$$ where $$ \alpha \in \mathbb{C}\!\,$$

Then from hypothesis we have for all $$z \in \{|z|=1\}\!\,$$,

$$ |f(z)| = |\alpha||g(z)|\leq |g(z)|\ \!\,$$

which implies

$$|\alpha| \leq 1 \!\,$$

Hence, by maximum modulus principle, for all $$z \in D \!\,$$

$$\left|\frac{f(z)}{g(z)}\right| = |\alpha| \leq 1 \!\,$$

i.e.

$$ |f(z)| \leq |g(z)|\!\,$$

Since $$0 \in D \!\,$$, we also have

$$ |f(0)| \leq |g(0)| \!\,$$