UMD Analysis Qualifying Exam/Jan09 Complex

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Problem 2[edit]

Solution 2[edit]

Problem 4[edit]

Suppose that f,g \!\, are analytic on \{ |z| \leq 1 \} \!\, with g \neq 0 \!\, on \{ |z| < 1 \} \!\,. Prove that  |f(z)| \leq |g(z)|\!\, for all z \in \{ |z|=1\} \!\, implies  |f(0)| \leq |g(0)|\!\,

Solution 4[edit]

Define new function h(z)[edit]

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

h is continuous on the closure of D[edit]

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[edit]

This follows from above

Case 1: h(z) non-constant on D[edit]

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[edit]

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)| \!\,

Problem 6[edit]

Solution 6[edit]