# Maxmimum\Minimum modulus principle

1. Feb 21, 2015

### MMS

Hey guys,

I'm studying for an exam that I have in Complex Analysis and I got stuck at the following question.

Question: True/False
f(z) is an analytical function in the ring 1 =< z =< 3. Also, the minimum of |f(z)| on |z|=1 equals to 1 and the maximum of |f(z)| on |z|=3 equals to 2.
Therefore, 1 =< |f(z)| =< 2 for every z in the ring 1 =< z =< 3.

the actual answer is false even though I don't get why I can't apply the maximum\minimum principle.

Here's my argument:

f is analytic in the ring and on its boundary (hence, bounded domain). Therefore, it is also continuous in the ring and up to its boundary. Then by the maximum modulus principle, it attains its maximum modulus on the boundary.
In other wording, |f(z)| on the boundary >= |f(z)| in the domain.
Same thing goes for the minimum modulus ---> |f(z)| on the boundary <= |f(z)| in the domain.
We're given that the maximum of the modulus is at |z|=3 and the minimum is at |z|=1. Hence, since those are the maximum and minimum of the modulus and it satisfies the principle,
1 =< |f(z)| =< 2 for every z in the ring 1 =< z =< 3.

What am I missing out here?

2. Feb 21, 2015

### Hawkeye18

First, you are not given maximum and minims on the boundary, you are given max on one part of the boundary and min on another part. So, maximum modulus principle does not apply.

Second, there is no "minimum modulus principle": consider function $f(z)=z$ on the unit disc $|z|<1$. Minimum of $|f(z)|$ of the circle $|z|=1$ is $1$, but $f(0)=0$.

Of course, to show that the statement is false, you need to construct a counterexample.

3. Feb 27, 2015

### mathwonk

just to get an idea, notice that for the function f(z) = z-2, |f(z)| has minimum equal to 1 on |z| = 1, and maximum equal to 5 on |z| = 3. but f(2) =0.

by the way do you suppose it changes anything if we assume also that f(z) is never zero?

Last edited: Feb 27, 2015
4. Feb 28, 2015

### Hawkeye18

Do you mean the original question? For the original question, even if we require that $f(z)$ is never zero, the statement is still false, so nothing changes.