# [SOLVED] Minimum of a complex function, proof

• Sep 16th 2009, 12:17 PM
arbolis
[SOLVED] Minimum of a complex function, proof
I've no idea what theorem can be useful to prove the following : Let $\displaystyle f$ be a continuous function in a closed and bounded region $\displaystyle R$, assume also that $\displaystyle f$ is analytic and not constant in the interior of $\displaystyle R$.
Assuming that $\displaystyle f(z)\neq 0$ in all $\displaystyle R$, prove that $\displaystyle |f(z)|$ has a minimum and that it is reached in the frontier of $\displaystyle R$ and never in the interior of $\displaystyle R$.
• Sep 16th 2009, 01:41 PM
Matt Westwood
I'm not sure, but I suspect this may be an application of Liouville's theorem:

http://www.proofwiki.org/wiki/Liouvi...mplex_Analysis)

... check it out anyway, there may be techniques and insights in there which will give you some inspiration.
• Sep 16th 2009, 02:41 PM
shawsend
If $\displaystyle f(R)\ne 0$, then what can you say about the (analytic) function $\displaystyle g(R)=\frac{1}{f(R)}$ using the Maximum Modulus Principle?
• Sep 16th 2009, 02:53 PM
arbolis
Quote:

Originally Posted by shawsend
If $\displaystyle f(R)\ne 0$, then what can you say about the (analytic) function $\displaystyle g(R)=\frac{1}{f(R)}$ using the Maximum Modulus Principle?

That $\displaystyle g$ reaches a minimum in the frontier of $\displaystyle R$.
Not sure I'm right. By the way I never heard this principle before. That's interesting, it reminds me of Extreme value theorem - Wikipedia, the free encyclopedia. Thanks a lot!
• Sep 16th 2009, 03:01 PM
shawsend
By the Maximum Modulus Principle, $\displaystyle g(z)$ reaches a maximum on the boundary which means that $\displaystyle f(z)$ must reach a minimum on the boundary as well.