Find the image of $f$ defined on $|z|<1$ complex unit disk, given by $$f(z)= \frac{1}{z}\prod_{k=1}^{n}(z-a_k)^{\lambda_k}$$where $|a_k|=1$, $0<\lambda_k<1$, $\sum_{k=1}^n\lambda_k=2$.

The first thing came to my mind is Schwarz-Christoffel integral $$S(z)=\int_0^z\frac{1}{\prod_{i=1}^n(\zeta-A_i)^{\beta_i}}d\zeta$$ which maps the real line onto a polygon. The image of $\infty$ under $S$ is not one of the vertices if $\sum_{k=1}^n\beta_k=2$.

I'm not sure how to relate the Schwarz-Christoffel integral to the $f$. Any suggestion is appreiciated.

  • $\begingroup$ I really doubt there is any nice characterization of the image. The boundary of the image should be contained in the image of the unit circle under $F$. By the way, you should specify which branch of $(z - \alpha_k)^{\lambda_k}$ you are using (presumably one where the branch cut is outside the unit circle). $\endgroup$ – Robert Israel Mar 11 '18 at 20:39
  • $\begingroup$ @Robert Israel I'm reading Stein, and I think he takes the branch which is positive when $z = x$ is real and $x > A_k$. $\endgroup$ – user136592 Mar 11 '18 at 20:53
  • $\begingroup$ If the $\lambda_k$'s are positive, then is the polygon no longer convex? $\endgroup$ – Jade Mar 12 '18 at 2:48
  • $\begingroup$ @Jade, it might be. One thing I notice is that the $a_i$ are sent to $0$, so the arcs between $ a_i$'s are sent to closed loops (simply by continuity). $0$ is sent to $infty$, thus I think the image should be the complex plane minus a flower-like closed region connected at $0$. $\endgroup$ – user136592 Mar 12 '18 at 2:53
  • $\begingroup$ So then the image is disjoint? But does that make sense? I thought a holomorphic function would send open regions to open regions. $\endgroup$ – Jade Mar 12 '18 at 2:59

Easy observations first:

  • The image is unbounded, since $0$ goes to $\infty$.
  • The points $a_1, \dots, a_n$ are mapped to $0$, so the image of boundary curve comes back to hit $0$ over and over
  • Each such return to $0$ makes the angle of $\pi \lambda_k$ with vertex $0$. The sum of these angles is $2\pi$.

Trying to imagine this leads to a conjecture: the image is the complement of the "star with no interior", i.e., the union of $n$ line segments joined at $0$.

To show that the guess is correct, it suffices to prove that $\arg f(z)$ remains constant on each arc between the points $a_k$. Branches may merit discussion elsewhere, but here all that matters is that we take some continuous branch of $\arg f$ on such an arc, and show it's constant.

As a warm-up, check that $$ \arg (1+z) = \frac12 \arg z $$ for every $z$ on the unit circle. Indeed, the triangle $-1, 0, z$ is isosceles, which implies its angle at $-1$ is $\frac12(\pi - (\text{angle at $0$}))$, which was the claim.

So, the rate of change of $\arg(1+z)$ is half the rate of change of $\arg z$. But this applies equally well to $\arg(z-a)$ for every unimodular $a$, since rotating the picture changes the arguments by a constant amount. It follows that the sum $$ -\arg z + \sum_{k=1}^{n} \lambda_k \arg(z-a_k) $$ has zero rate of change, thanks to the condition $\sum \lambda_k = 2$. And this was $\arg f(z)$.


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