Ahlfors Complex Integration

Question

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This is my opinion on the question. Is true or not? If not what is the useful
solution? Which way is more useful?

I wouldn't call this an "opinion", but rather an attempt at solving the problem! — Patrick Da Silva, Jun 21 '14 at 16:19
:) thank u Patrick Da silva. I love to attempt good questions. but maybe it is easy for u or other people. I got very bad education from undergraduate unfortunately and that why i am happy when i solve the problem myself — Analysis, Jun 21 '14 at 16:23
No I mean your attempt is perfectly fine, I was only commenting the choice of words! Don't worry, it's all good :) — Patrick Da Silva, Jun 21 '14 at 16:28

score 0 · Answer 1 · answered Jun 21 '14 at 16:11

0

It follows from the original inequality that the logarithm of $f(z)$ is single-valued in the domain $\Omega$ and hence the integral of its differential along any closed curve is zero.

answered Jun 21 '14 at 16:11

Vladimir

5,702

My opinion is wrong? or not Professor – Analysis Jun 21 '14 at 16:12
No it is not wrong, you actually do the same thing without mentioning the logarithm. – Vladimir Jun 21 '14 at 16:14
Ohh thank you i am feeling happy that i solved myself :) – Analysis Jun 21 '14 at 16:15

score 0 · Answer 2 · answered Jun 22 '14 at 14:25

Let $\mu(t) = f\circ\gamma(t)$. Then $$ \int_{\mu}\frac{dw}{w} = \int_{\gamma}\frac{f'(z)}{f(z)}\,dz\;. $$ The curve $\mu$ satisfies $|\mu(w)-1| < 1$, which means that the curve $\mu$ lies in the open unit ball centered at $1$. In this region, $1/w$ has a holomorphic antiderivative and, so, the integral on the far left is $0$.

Ahlfors Complex Integration

2 Answers2