# [SOLVED] Minimum of a complex function, proof

• September 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 $f$ be a continuous function in a closed and bounded region $R$, assume also that $f$ is analytic and not constant in the interior of $R$.
Assuming that $f(z)\neq 0$ in all $R$, prove that $|f(z)|$ has a minimum and that it is reached in the frontier of $R$ and never in the interior of $R$.
• September 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.
• September 16th 2009, 02:41 PM
shawsend
If $f(R)\ne 0$, then what can you say about the (analytic) function $g(R)=\frac{1}{f(R)}$ using the Maximum Modulus Principle?
• September 16th 2009, 02:53 PM
arbolis
Quote:

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

That $g$ reaches a minimum in the frontier of $R$.
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!
• September 16th 2009, 03:01 PM
shawsend
By the Maximum Modulus Principle, $g(z)$ reaches a maximum on the boundary which means that $f(z)$ must reach a minimum on the boundary as well.