Author Topic: Complex Analysis Questions (Read 1848 times) Tweet Share

Gloamglozer · « **on:** August 27, 2011, 03:46:35 pm »

Please note that the below two questions are assignment questions. Please feel free to reveal as much or as little as you desire. Thank you.

Let $f(z)=u(x,y) + iv(x,y)$ be a complex function. Consider a general change of variables $x( \eta, \phi )$ and $y( \eta, \phi )$ producing the function:

$\~{f}( \eta, \phi) = f(x(\eta,\phi),y(\eta,\phi))$

1. Consider the change of variables $\eta=x+iy$ and $\phi=x-iy$ and show that

$\frac{\partial \~{f}}{\partial \eta} = \frac{1}{2} (\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}) + \frac{i}{2} (\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y})$

$\frac{\partial \~{f}}{\partial \phi} = \frac{1}{2} (\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y}) + \frac{i}{2} (\frac{\partial v}{\partial x} + \frac{\partial u}{\partial y})$

2. Explain why this shows that an analytic function is independent of $\bar{z}$ .

EDIT: Post edited to put tildes in.

kamil9876 · « **Reply #1 on:** August 27, 2011, 08:20:53 pm »

that looks pretty confusing because $f$ was used to define two different things, I am assuming that we are using the "new" $f$ in the partial derivatives, that's why I advise you to firstly write $g(\eta,\phi) = f(x(\eta,\phi),y(\eta,\phi))$ . Then you should be able to use the chain rule to find $\partial g /\partial \eta$ etc.

Who's your lecturer?

Gloamglozer · « **Reply #2 on:** August 27, 2011, 09:38:32 pm »

Ahh ooops. There were supposed to be tildes on the partial derivatives of $\partial f /\partial \eta$ etc.

Iwan Jensen is the lecturer this semester. Last semester they had Alex Ghitza/Paul Norbury. I think it was supposed to be Paul Pearce because we're using his notes as a backbone but for some reason Iwan was drafted in at short notice.

Gloamglozer · « **Reply #3 on:** October 18, 2011, 10:15:57 pm »

Suppose $f(z)$ is an entire function such that $|f'(z)| \leq |z|$ $\forall z$ . Show that $f(z) = a + bz^2$ with $|b| \leq 1$ .

humph · « **Reply #4 on:** October 18, 2011, 10:32:28 pm »

It suffices to show that if $f(z)$ is an entire function with $|f(z)| \leq z$ for all $z \in \mathbb{C}$ , then $f(z) = bz$ for some $|b| \leq 1$ . The key is that $f(z)$ being entire means that
$f(z) = \sum^{\infty}_{n=0}{a_n z^n}$
with
$a_n = \frac{1}{2\pi i} \oint_{\partial B_r(0)}{\frac{f(z)}{z^n} \, dz}$
for all $r > 0$ . As you can choose $r$ as big or as small as you like, you can estimate simply and show that $a_n = 0$ unless $n = 1$ , in which case $|a_1| \leq 1$ .

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