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the question is from Conway's Functions of One Complex Variable, volume I,second edition, chapter VI section 1, exercise 7.

Let $f$ be analytic in the disk $B(0,R)$ and for $0 \leq r \leq R$ define $$A(r)=\max\{\operatorname{Re} f(z) : |z|=r\}.$$

Show that unless $f$ is constant, $A(r)$ is strictly increasing function of $r$.

Now obviously from the maximum modulus we must have for any $r_1< r_2$ and $|z|=r_1$,$|\zeta|=r_2$, $|f(z)|\geq |f(\zeta)|\geq \operatorname{Re} f(\zeta)$, but don't see how use for the real parts here.

Only hints if you can.

Thanks.

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Hint: Adapt the proof of the maximum modulus principle from the open mapping theorem. –  t.b. Oct 13 '11 at 21:07

1 Answer 1

up vote 4 down vote accepted

Hint: consider $g(z)=e^{f(z)}$.

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Thanks, so easy, I need to remember this trick. –  MathematicalPhysicist Oct 14 '11 at 6:31
    
Well, I do not understand what is going one here could any one tell me about the solution? –  Une Femme Douce Apr 30 '13 at 12:31
    
@UneFemmeDouce $|e^{f(z)}|=e^{\Re f(z)}\le e^{A(r)}$ and maximum modulus principle on $|z|=r$, $|z|=s$, $r<s$. –  felipeuni Aug 25 at 4:12

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