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.
Best Answer
You are thinking too complicated. $\Re \bigl(\overline{z}f(z)\bigr) > 0$ on the unit circle implies that the function
$$g_\lambda(z) = (1-\lambda)z + \lambda f(z)$$
has no zeros on the unit circle for $\lambda \in [0,1]$:
$$\Re \bigl( \overline{z}g_\lambda(z)\bigr) = (1-\lambda)\lvert z\rvert^2 + \lambda \Re \bigl(\overline{z}f(z)\bigr) > 0$$
for $\lvert z\rvert = 1$. The number of zeros of $g_\lambda$ in the unit disk is therefore
$$N(\lambda) = \frac{1}{2\pi i} \int_{\lvert z\rvert = 1} \frac{g_\lambda'(z)}{g_\lambda(z)}\,dz = \frac{1}{2\pi i}\int_{\lvert z\rvert = 1} \frac{(1-\lambda) + \lambda f'(z)}{(1-\lambda)z + \lambda f(z)}\,dz.$$
Since $g_0(z) = z$, we know $N(0) = 1$. We also know that $N(\lambda)$ depends continuously on $\lambda$.
Note that the argument is essentially Rouché's theorem. The condition $\lvert f-g\rvert < \lvert f\rvert$ in Rouché's theorem is just a simple way of stating a condition that ensures that $\lambda f(z) + (1-\lambda)g(z)$ has no zeros on the contour. $\Re \bigl(\overline{z}f(z)\bigr) > 0$ on $\lvert z\rvert = 1$ is another such condition.