Check that $f(0)=\frac{1}{2}$ using Schwarz Lemma Suppose that $f$ is analytic in a regoin containing the closure of $D=\{|z|<1\}$. Assume that $|f(z)|<1$ for $z\in D$.
Assume that $f$ has a simple zero at $\frac{1}{4}(1+i)$ and a double zero at $\frac{1}{2}$.
Can $f(0)=\frac{1}{2}$?
Since $|f(z)|<1$, I think we can use the Schwarz Lemma by applying automorphism $\phi_a$ where $$\phi_a(z)=\frac{z-a}{1-\bar{a}z}$$
Here I tried $g=f\phi_{-\frac{1}{4}(1+i)}$ and $g=f\phi_{-\frac{1}{2}}$ but I still can't get enough information to verify whether $f(0)=\frac{1}{2}$.
Even I tried to write $$f(z)=(z-\frac{1}{4}(1+i))(z-\frac{1}{2})^2h(z)$$
for some $h$ analytic and $h(\frac{1}{2})\neq 0$ and $h(\frac{1}{4}(1+i))\neq 0$, still it can't work.
Remark:
This is Exercise 7 (Page 133) from Function of One Complex Variable by Conway.
There is a solution in http://sertoz.bilkent.edu.tr/courses/math503/2014/hwk03-sol.pdf. But I am not satisfied with it since we cannot conclude that $|f(z)|=1$ for $|z|=1$. 
 A: Claim: It is impossible for $f(0)=1/2$.
Proof. Suppose, for contradiction, that $f(0)=1/2$. Thanks to the fact that $f:\mathbb{D}\to\mathbb{D}$ is analytic, and that
$$
\Phi_{a,\theta}(z)=e^{i\theta}\frac{z-a}{1-\bar{a}z}
$$
is analytic from $\mathbb{D}$ to $\mathbb{D}$ with $\Phi_{a,\theta}(a)=0$, define
$$
g(z)=\left(\Phi_{\frac{1}{2},\theta}\circ f\right)(z)=e^{i\theta}\frac{f(z)-\frac{1}{2}}{1-\frac{1}{2}f(z)},
$$
which satisfies that $g:\mathbb{D}\to\mathbb{D}$ is analytic with $g(0)=0$. So Schwarz lemma applies:
$$
\left|g(z)\right|\le\left|z\right|\quad\forall z\in\mathbb{D}.
$$
Note that $1/2\in\mathbb{D}$, and that, using $f(1/2)=0$,
$$
\left|g\Bigl(\frac{1}{2}\Bigr)\right|=\left|e^{i\theta}\frac{0-\frac{1}{2}}{1-\frac{1}{2}\cdot 0}\right|=\left|\frac{1}{2}\right|.
$$
Thus by Schwarz lemma, it is a must that
$$
g(z)=e^{i\alpha}z\quad\forall z\in\mathbb{D}
$$
for some real $\alpha$. Combined with the definition of $g$ from above,
$$
f(z)=\frac{\frac{1}{2}+e^{i\left(\alpha-\theta\right)}z}{1+\frac{1}{2}e^{i\left(\alpha-\theta\right)}z}.
$$
Further, $f(1/2)=0$ forces $\alpha-\theta\equiv\pi$, i.e.,
$$
f(z)=\frac{\frac{1}{2}-z}{1-\frac{1}{2}z}=\frac{2z-1}{z-2}.
$$
One may see that this only possibility for $f$ yields neither a simple zero at $z=\left(1+i\right)/4$ nor a double zero at $z=1/2$. Therefore, $f(0)\ne 1/2$.
