Show that the given function has one zero inside the unit disk

Question

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?

I am just curious: Did you manage to solve https://math.stackexchange.com/q/4165877/42969? — Martin R, Jun 15 '21 at 09:38
@MartinR I just deleted it because there is a possibility that the question itself is wrong. No one actually answered the question for a while. But if you want I'll undelete it. — one potato two potato, Jun 15 '21 at 12:31
I don't think that is necessary (unless you know definitely that the question is correct). Thanks. — Martin R, Jun 15 '21 at 13:33

Martin R · Accepted Answer · 2021-06-10T14:05:31.217

4

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 $$ 0 = \frac{1}{2\pi i}\int_\Gamma \frac{dw}{w} = \frac{1}{2\pi i} \int_\gamma \frac{g'(z)}{g(z)} \, dz \\ = \frac{1}{2\pi i} \int_\gamma\left(\frac{f'(z)}{f(z)} -\frac 1z\right) \, dz = -1 + \frac{1}{2\pi i} \int_\gamma \frac{f'(z)}{f(z)} \, dz \, . $$

edited Jun 10 '21 at 14:05

answered Jun 10 '21 at 13:02

Martin R

113,040

That's an unexpected nice solution. But unfortunately I don't know that. I only know classical form of Rouche's theorem. – one potato two potato Jun 10 '21 at 13:10
Can I easily derive that symmetric form from the original version? – one potato two potato Jun 10 '21 at 13:11
1

@barista: I have added another argument (that is essentially the proof of Rouché's theorem). – Martin R Jun 10 '21 at 13:15
Wait, I think you're using fundamental theorem of integration. But to use that $\log\frac{f(z)}{z}$ should be continuous on some open set containing $|z|\leq 1$ but I wonder if that's true because function is not defined on $z=0$ – one potato two potato Jun 10 '21 at 13:35
@barista: I have added a more detailed calculation, does that make it clear? – Martin R Jun 10 '21 at 13:45
Why the first $0=$ true? Assumption of Cauchy' theorem is on the simply connected domain of the integrand. – one potato two potato Jun 10 '21 at 13:53
@barista: $\Gamma$ is a curve in the right half-plane, which is simply connected. – Martin R Jun 10 '21 at 13:54
Yes but it's the image of the integrand not the domain of it – one potato two potato Jun 10 '21 at 13:55
Let us continue this discussion in chat. – Martin R Jun 10 '21 at 13:56

Show that the given function has one zero inside the unit disk

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