(New page: Post solutions for mock qual #2 here. Please indicate authorship! ----) |
|||
| Line 1: | Line 1: | ||
| − | Post solutions for mock qual #2 here. Please indicate authorship! | + | Post solutions for mock qual #2 here. Please indicate authorship! |
---- | ---- | ||
| + | |||
| + | <br> | ||
| + | |||
| + | == Problem 1 == | ||
| + | |||
| + | == Problem 2 == | ||
| + | |||
| + | Suppose <math>u: \mathbb C \to \mathbb R</math> is a non-constant harmonic function. Show that the zero set <math> | ||
| + | S = \{z \in \mathbb C | u(z) = 0\} | ||
| + | </math> is unbounded as a subset of <math>\mathbb C</math>. | ||
| + | |||
| + | === Clinton, 2014 === | ||
| + | |||
| + | Suppose for contradiction that <span class="texhtml">''S''</span> is bounded; that is, <math>\exists R \forall z</math> such that <math>|z| \geq R \implies u(z) \neq 0</math>. Let <span class="texhtml">''f'' = ''u'' + ''i''''v''</span> analytic, where <span class="texhtml">''v''</span> is the global analytic conjugate for $u$. We will show that <span class="texhtml">''g'' = ''e''<sup>''f''(''z'')</sup></span> is constant, thus $u$ is constant. | ||
| + | |||
| + | <br> | ||
| + | |||
| + | == Problem 3 == | ||
| + | |||
| + | == Problem 4 == | ||
| + | |||
| + | == Problem 5 == | ||
| + | |||
| + | == Problem 6 == | ||
| + | |||
| + | == Problem 7 == | ||
Revision as of 06:27, 7 August 2014
Post solutions for mock qual #2 here. Please indicate authorship!
Contents
Problem 1
Problem 2
Suppose $ u: \mathbb C \to \mathbb R $ is a non-constant harmonic function. Show that the zero set $ S = \{z \in \mathbb C | u(z) = 0\} $ is unbounded as a subset of $ \mathbb C $.
Clinton, 2014
Suppose for contradiction that S is bounded; that is, $ \exists R \forall z $ such that $ |z| \geq R \implies u(z) \neq 0 $. Let f = u + i'v analytic, where v is the global analytic conjugate for $u$. We will show that g = ef(z) is constant, thus $u$ is constant.
