Let $f$ be a holomorphic function in a neighborhood of the closed unit disc $\{z \in \mathbb{C} : |z| \leq 1\}$, and suppose that $\Re{(\bar{z}f(z))} > 0 $ when $|z| = 1$. Prove that $f$ has exactly one zero in the open unit disc.
Till now what I have tried is this:
When $|z| = 1: \quad \Re{(\bar{z}f(z))} > 0 \implies |\bar{z} f(z)| > 0 \implies |f(z)| > 0 $. Since $|z|=1$ is compact, $|f(z)| \leq M $ on $|z|=1$. WLOG, we can assume $M=1$. This means that $f(z)$ is zero-free on the unit circle and thus has only finitely many zeros inside the unit disc. Thus $f(z)$ has the following Blaschke product representation -
$f(z) = z^k \prod_{j=1}^{n} \dfrac{-a_j}{|a_j|} \dfrac{z-a_j}{1-\bar{a_j}z}F(z)$ where $f(a_j) = 0$ for all $j$ and $F(z)$ is a bounded, zero-free holomorphic function in the unit disc. I do not know how to proceed from here and if this is at all useful. Can somebody help please.