Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
Hint: Adapt the proof of the maximum modulus principle from the open mapping theorem. – t.b. Oct 13 '11 at 21:07
up vote 4 down vote accepted

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

share|cite|improve this answer
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? – Un Chien Andalou 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 '14 at 4:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.