Image of a function under unit disk.

Question

What can we say about the image of the following function under open unit disk: $$f(z)=\frac{1}{(1-z)(1-a z)},\quad 0<a\leq1.$$

I think the complement of the image domain is a convex set. But I don't have proof.

Answer 1

Let, $$w=\frac{1}{(1-z)(1-az)}$$

$$z=\frac{(a+1)\pm \sqrt{(a+1)^2-4a(1-1/w)}}{2a}$$

Now, $|z|<1$ implies $$2a>\left|(a+1)\pm \sqrt{(a+1)^2-4a(1-1/w)}\right|>(a+1)-\left|\sqrt{(a+1)^2-4a(1-1/w)}\right|$$

$$\implies (a-1)^2>|(a+1)^2-4a(1-1/w)|>(a+1)^2-4a|1-1/w|$$

$$\implies |1-1/w|>1\implies |1-w|^2>|w|^2$$

$$(1-w)(1-\bar w)>|w|^2\implies \frac{w+\bar w}{2}<1/2\implies \Re (w)<1/2$$

So, image of the open unit disc under the map $f$ is $\{z:\Re(z)<1/2\}$.

Answer 2

Whatever you have done is correct. This shows that the image domain is contained in $\{z: \Re (z)<1/2 \}$. But I want exactly the image domain or some geometric description about the image domain (such as convexity, concavity).