Find the image of unit disk under $f(z)= \frac{1}{z}\prod_{k=1}^{n}(z-a_k)^{\lambda_k}$ 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.
 A: 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 $\arg f(z)$.
