4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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 $\arg f(z)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.