# Thread: [SOLVED] Minimum of a complex function, proof

1. ## [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 $R$.

2. 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.

3. 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?

4. 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!

5. 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.