Möbius transformation from Im(z)>0 circle

Question

My question is:

Let $f(z)=\frac{2z+i}{2+iz}$ and $D^+=\left \{ z\in\mathbb{C} : \left | z \right |<1, Im(z)>0 \right \}$.

Find $f(D^+)\subseteq \bar{\mathbb{C}}$ and draw it.

I know that given a set of three distinct points $z_1, z_2, z_3$ on the Riemann sphere and a second set of distinct points $w_1, w_2, w_3$, there exists precisely one Möbius transformation.

I know also that $f(D^+)$ is not a line because $z$ has to be $2i$ and it contradicts $D^+$ definition.

I thought about finding 3 points from $D^+$ but I can't see how I can form $f(D^+)$ from it.

Any hint or help would be appreciated.

Answer 1

Recall that a Mobius transform maps (part of) a circle to either (part of) a circle or a straight line (segment), and maps (part of) a straight line to either (part of) a circle or a straight line (segment) as well. Therefore, it suffices to figure out what $f$ maps the upper-half circle and the line segment $\left(-1,1\right)$ to. These would imply the boundary of $f(D^+)$.

Note that

• $f(-1)=-1$, and that
• $f(0)=i/2$, and that
• $f(1)=1$.

These facts imply that $f$ maps $(−1,1)$ to part of the circle (i.e., an arc) passing through $-1$, $i/2$ and $1$, with $-1$ and $1$ being the end points of this arc.

Likewise, note that

• $f(1)=1$, and that
• $f(i)=3i$, and that
• $f(-1)=-1$.

These facts imply that $f$ maps the upper-half circle to part of the circle (again, an arc) passing through $1$, $3i$ and $-1$, with $1$ and $-1$ being its end points.

Further, note that

• $i/2\in D^+$, and that
• $f(i/2)=4i/3$.

These facts imply that $f$ maps $D^+$ to the interior of the domain determined by the two arcs depicted from above.

Consequently, the $f(D^+)$ should be as follows.