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I have the following question from my Complex Analysis Class:

$f(z)$ is analytic in the unit disk $D=\{z:|z|<1\}$ and continuous in the disk $\bar{D}=\{z:|z|\leq 1\}.$ Suppose also $f(z)/z^3$ can be extended to be analytic in all of $D$ (including the origin). If $|f(z)|\leq 2 in D$, what is the maximum value that $|f(0.4+0.5i)|$ can assume under those conditions?

This problem is really throwing me off because I have all of these theorems and examples that kind of hint at ways of forming a solution, however none of them work perfectly for this example. I think what's throwing me off is the fact that we are talking about our function in terms of $f(z)$ without an actual function given.

I know that the function will reach its max on its boundary, $|z|=1$, which can be parametrized as $z=e^{it}$, $0\leq t \leq 2\pi$. But then there really isn't anywhere to exactly plug this in.

Any help is much appreciated!

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Hint: Since $g(z)=f(z)/z^3$ extends to an analytic function, you must also have $|g(z)|\leq 2$ in $D$. So $|f(z)|\leq 2|z|^3$ (optimal bound) and you may calculate the value at the chosen point.

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    Ohhhh wow I didn't even think of this. Ok, so is this the point where I can use my $z=e^{it}$ to see that $|f(z)|\leq 2\cdot e^{3it}$ ? – 2017-02-13
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    yes, you can do so since $f$ is also assumed to be continuous on the boundary (if not you could look at a circle of slightly smaller radius and take a limit). Note, however, that you should look at the abs value on the RHS (which then becomes 2). – 2017-02-13
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    This was a great answer, thanks for your help! – 2017-02-13