Let $f$ be an analytic function on $\{z:|z|\leq 1\}$ such that $\operatorname{Re}(\overline{z}f(z))>0$ for $|z| =1$. Then $f$ has one simple zero on the unit disk.
My attempt: I tried to show this using argument principle i.e. Claim : $\frac{1}{2\pi i}\int_{|z|=1}\frac{f'(z)}{f(z)}=1$. To see this, $\frac{1}{2\pi i}\int_{0}^{2\pi}\frac{f'(e^{i\theta})}{f(e^{i\theta})}ie^{i\theta}d\theta = \frac{1}{2\pi}\int_0^{2\pi}\frac{f'(e^{i\theta})}{f(e^{i\theta})e^{-i\theta}}d\theta$. I got $f(z)\overline{z}$ on the denominator but I don't know how to apply the given condition since even if I take the real part of the integral, I don't have any information of $f'$. Could you give any hints?
Best Answer
One can apply Rouché's theorem (in its symmetric form): For $|z|=1$ is $$ |f(z)-z|^2 = |f(z)^2| - 2\operatorname{Re}(\overline{z}f(z)) + |z|^2 < |f(z)^2|+ |z|^2 \le (|f(z)| + |z|)^2 $$ and therefore $|f(z)-z| < |f(z)| + |z|$. This implies that $f(z)$ and $z$ have the same number of zeros in the unit disk.
Alternatively you can show that $$ \frac{1}{2\pi i}\int_{|z|=1}\left(\frac{f'(z)}{f(z)} -\frac 1z\right) \, dz $$ is zero because the integrand is the derivative of $$ \log \frac{f(z)}{z} $$ which is holomorphic because $f(z)/z$ lies in the right half-plane.
More detailed: Let $g(z) = f(z)/z$, $\gamma(t) = e^{it}$ for $0 \le t \le 2\pi$, and $\Gamma = g \circ \gamma$. $\Gamma$ lies in the right half-plane where the principal branch of the logarithm is holomorphic, with $(\log w)' = 1/w$. It follows that $$ 0 = \frac{1}{2\pi i}\int_\Gamma \frac{dw}{w} = \frac{1}{2\pi i} \int_\gamma \frac{g'(z)}{g(z)} \, dz \\ = \frac{1}{2\pi i} \int_\gamma\left(\frac{f'(z)}{f(z)} -\frac 1z\right) \, dz = -1 + \frac{1}{2\pi i} \int_\gamma \frac{f'(z)}{f(z)} \, dz \, . $$