Find $|f(0)|$ given $|f(z)|$ on boundary of unit disk 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. 
 A: 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 $|f(0)|=e^{1/6}$
A: 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!
