What is the image of D under the map $f(z)=bz+\frac{1}{z}$?

Question

Let $f:\mathbb C_\infty\to \mathbb C_\infty$ be a sum of Mobius transformations defined by $f(z)=bz+\frac{1}{z}$, where $-1<b<1$.

I've succeeded to find the image of the unit closed disc $D$ under the both of Mobius transformations $f_1(z)=bz$ and $f_2(z)=\frac{1}{z}$ individually.

But, I failed to find $f(D)$.

Question : Find $f(D)$. Is it true that $f(D)$ is a subset of $\mathbb C-D$?

Answer 1

Counter example

For $-1<b<0$, consider$$z_0={1\over \sqrt {1-b}}\implies |z_0|<1$$therefore$$f(z_0)=bz_0+{1\over z_0}=z_0$$

P.S.

This transformation maps the circle $$x^2+y^2=r^2$$to the ellipse $${x^2\over b+{1\over r^2}}+{y^2\over b-{1\over r^2}}=r^2$$