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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?

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# Maxmimum\Minimum modulus principle

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