$f$ holomorphic on unit disk, $|z|=1\implies|f(z)|=1$, $f(\frac12)=f(-\frac12)=0$ 
Find a holomorphic function $f$ on $B(0,1)$, continuous on $\overline{B(0,1)}$, such that $|z|=1\implies|f(z)|=1$, and furthermore $f(\frac12)=f(-\frac12)=0$.

My approach was to look at the function $h_a(z)=\frac{z-a}{1-\overline{a}z}$, for $|a|<1$, which clearly satisfies the first properties. I was motivated by a theorem stating that every biholomorphic function from the unit disk onto itself are of this form $c\frac{z-a}{1-\overline{a}z}$, for $c\in C(0,1)$. However, since we need two zeroes, a biholomorphic function would surely not work here. For example $h_{1/2}(z)(z+\frac12)$ does also not work, since it contradicts the second required property; similarly for $h_{1/2}(z)(z+\frac12)/|z+\frac12|$. Also a linear combination $th_{1/2}+(1-t)h_{-1/2}$ seems to contradict the second requirement.
My question: am I on the right track? If yes, any hints how to proceed? If not, would you know any alternative approach?

You could consider the function $$f(z) = \bigg(\frac{z+1/2}{1+1/2z}\bigg) \bigg(\frac{z-1/2}{1-1/2z}\bigg) = h_{1/2}(z)*h_{-1/2}(z).$$
You already know that if $|z|=1$ then $|f(z)| = |h_{1/2}(z)||h_{-1/2}(z)| = 1.$
The function $f$ defined this way is called a finite Blaschke product. In fact, all the functions that verify your conditions are of the form $$\phi(z) = k\bigg(\frac{z+1/2}{1+1/2z}\bigg)^{m_1} \bigg(\frac{z-1/2}{1-1/2z}\bigg)^{m_2},$$ with $m_1, m_2 > 0$ and $|k|=1.$
