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Suppose that $f$ is a non-constant, smooth, complex valued function on $\mathbb{C}$ with $\Gamma=\{z\in \mathbb{C}: \lvert f(z)\rvert=7\}$ a smooth, closed, simple curve. Let $G$ be the enclosed domain and suppose that $f$ is analytic on $G$.

It is easy to show using the maximum principle that $f$ must have at least one zero in $G$. Now suppose also that $f'\equiv \frac{\partial f}{\partial z}$ has no zeroes on $\Gamma$ and that $f$ has $m$ zeroes in $G$ counting multiplicity. How many zeroes must $f'$ have?

I think that the answer has to be $m-1$. My approach was to try to use the argument principle. I don't exactly see how to get a handle on $\int_{\Gamma} \frac{f''}{f'}dz$. Next I thought about parameterizing $\Gamma$ as $\gamma(t)$ for $a\leq t\leq b$ with $\lvert \gamma'(t)\rvert =1$ and trying to use $f'(\gamma(t))=\frac{\frac{d}{dt}f(\gamma(t))}{\gamma'(t)}$, then computing the change in argument around $\Gamma$. This sort of made sense to me as I'm pretty convinced that the change in argument of $\frac{d}{dt}f(\gamma(t))$ is the same as the change in argument of $f(\gamma(t))$, but I couldn't prove that and even with this fact the proof is quite hand wavy.

Anyway if someone has a more elegant approach, please share!

MSA2016
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  • We can note that if the answer indeed depends only on $m$ and not on $f$, then the answer must be $m-1$, as the example $f(z)=z^m$ shows. – Greg Martin Jul 13 '21 at 00:52
  • I thought about that, but I'm not sure how to do that without changing the fact that the modulus is constant on the boundary. One workaround might be first passing through a map conformal map $\zeta: D(0,1)\rightarrow G$ which we know exists via the Riemann Mapping Theorem. Then $\frac{1}{7}f\circ \zeta$ is a finite Blaschke product. You can then divide out a Blaschke factor and not change the value of the modulus on the boundary. – MSA2016 Jul 13 '21 at 02:01
  • Sorry for the stupid comment, induction clearly doesn't work. Yes, at this point it's pretty clear that you have to take more advantage of the fact that the domain is very regular, than just the argument principle applying. – Sarvesh Ravichandran Iyer Jul 13 '21 at 02:02
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    I think that the condition on the derivative not vanishing on $\Gamma$ is superfluous since that follows from the fact that $\Gamma$ is a simple closed curve and the fact that (finite) Blaschke products never have critical points on the boundary, since as noted above $B=\frac{1}{7}f\circ \zeta$ is such and the zeroes of $B'$ in $\mathbb D$ and on the unit circle are the zeroes of $f'$ in $G$ and on $\Gamma$ respectively since $\zeta$ is a Riemann map; by general Blaschke theory it follows that $B'$ has $m-1$ zeroes inside the unit disc, hence $f'$ has $m-1$ zeroes inside $G$ as expected – Conrad Jul 13 '21 at 03:43
  • the result about critical points of finite Blaschke products $B$ is very easy to prove when $B$ has distinct zeroes and $B(0)\ne 0, B'(0) \ne 0$ and then it follows from the fact that any Blaschke product of degree $n$ is uniformly approximated inside the closed unit disc by Blaschke products of degree $n$ with simple zeroes and which do not vanish at the origin, nor have criticcal points there – Conrad Jul 13 '21 at 03:48
  • Unless I am mistaken, this follows from the Riemann-Hurwitz formula since $f$ is a proper map between simply connected domains. – Martin R Jul 13 '21 at 05:19

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