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This is from a previous qual exam in Analysis:

Suppose $f(z)$ is a non-vanishing analytic function on $D=\{z: |z|<1\}$ and continuous on $\overline{D}$. Suppose that $|f(e^{2\pi it})|=e^{t(1-t)}$ for $t \in [0,1]$, find $|f(0)|$.

I have tried the maximum and minimum modulus principles, but both only give an inequality for $|f(0)|$, namely, $1 < |f(0)|<e^{\frac{1}{4}}$. I have no clue how to proceed from there.

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Because $f(z)$ is non-zero in $D$ (which is simple-connected), the log of $f(z)$ is well defined and analytic in $D$. Define $g(z):=\ln(f(z)) = \ln(|f(z)|)+i\phi(z)$. We apply Cauchy's theorem to $g(z)$ to get $g(0)=\ln(|f(0)|)+i\phi(0)$:

$$ \begin{align} g(0) = \ln(|f(0)|)+i\phi(0) &= \frac{1}{2\pi i}\int_{\partial D}\left(\frac{\ln(|f(z)|)}{z}+i\frac{\phi(z)}{z}\right)\,dz \\ &=\int_0^1t(1-t)\,dt+i\int_0^1\phi(e^{2\pi it})dt. \end{align}$$

The real part of this equation gives:

$$ \ln(|f(0)|) = \int_0^1t(1-t)\,dt = \frac{1}{6}.$$

So $|f(0)|=e^{1/6}$

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  • $\begingroup$ Your solution solves my problem perfectly. Thanks a lot! $\endgroup$ – Yuxin Wang Aug 2 '17 at 19:59
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I haven't thought about this very long, and this probably won't work, but the statement of the problem leads me to suspect that the conditions are strong enough to argue that $g(z)=|f(z)|$ is analytic inside the disk. If this is true, you can use the Cauchy integral formula to get $g(0)=|f(0)|$. I think that the result would be something like $$ |f(0)| = \int_0^1 e^{t(1-t)} dt $$ which should be expressible using $\text{erf}(z)$. Of course, this is predicated on a huge assumption that I haven't bothered to think through!

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  • $\begingroup$ $|f(z)|$ is real and non constant, so cannot be analytic. $\endgroup$ – ajotatxe Aug 2 '17 at 19:30
  • $\begingroup$ You're right. I guess I said that it probably wouldn't work. But it seems that they are setting you up to use the Cauchy integral formula. Perhaps it should be applied to another function that is analytic and indirectly shows the same result, perhaps something like $$f(z) + \overline{f(\bar{z})}$$. $\endgroup$ – sasquires Aug 2 '17 at 19:35
  • $\begingroup$ That one is also real :) $\endgroup$ – ajotatxe Aug 2 '17 at 22:12

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