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Suppose that $f$ is analytic in the annulus: $1 \leq \vert z \vert \leq 2 $, that $\vert f \vert \leq 1$ for $\vert z \vert = 1$ and that $\vert f \vert \leq 4$ for $\vert z \vert = 2$. Prove $\vert f(z) \vert \leq \vert z \vert ^2$ throughout the annulus.

I know that I would have to apply the Maximum Modulus Theorem here, but I am having trouble figuring out how to do so. Would I have to use the analyticity of $f$ in order to reach such a conclusion?

I am using the textbook Complex Analysis, Third Edition by Joseph Bak and Donald J. Newman.

Any suggestions and tips are greatly welcomed.

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What can you say about $f(z)/z^2$ on the boundary of the annulus? –  WimC Apr 7 '13 at 6:12
    
If $z =1$, then it the inequality holds. If $z=2$, the inequality is still true. The Maximum Modulus Theorem states that a non-constant analytic function in a region $D$ does not have any interior maximum points. Using $\frac {f(z)}{z^2}$, then it does not have any interior maximum points; hence, it assumes its maximum modulus at its boundary points? –  Jamil_V Apr 7 '13 at 6:28
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@Jamil_V Why not post your arguments as an answer, so that we know what you already have, and that we can close the issue? Regards. –  awllower Apr 7 '13 at 7:14

1 Answer 1

As suggested, my comment above will be posted as an answer to seek clarification. If $z=1$, then it the inequality holds. If $z=2$, the inequality is still true. The Maximum Modulus Theorem states a non-constant analytic function in a region $D$ does not have any interior maximum points. Using $\frac{f(z)}{z^2}$, then it does not have any interior maximum points; hence, it assumes its maximum modulus at its boundary points?

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I think it is clear that, if $f$ is constant, then the inequality trivially holds. So, in any case, that inequality is true, and the proof is complete.Regards. –  awllower Apr 9 '13 at 0:36

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