# Möbius transformation from Im(z)>0 circle

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.

• Mobius transform maps (part of) a circle to (part of) a circle or a straight line, and maps (part of) a straight line to (part of) a circle or a straight line as well. So you need to figure out what your transform maps the upper-half circle and the $\left(-1,1\right)$ interval to. Apr 15, 2018 at 9:49
• Do I have to take 3 point on the boundry? Something like $z_1=1,z_2=-1,z_3=i$ because it doesn't belong to $D^+$. Apr 15, 2018 at 10:16
• Points on the boundary are fine. Or you may substitute $e^{i\theta}$ with $\theta\in\left(0,\pi\right)$ and $x$ with $x\in\left(-1,1\right)$ into the transform, and see what the transform maps these curves to. As you may have been aware, Mobius transform maps the boundary of some open domain to the boundary of its image. So getting the profile of the boundaries suffices to tell you the image. Apr 15, 2018 at 10:47
• Alternatively, note that $f(1)=1$, $f(-1)=-1$, $f(0)=i/2$, for which $f$ maps $\left(-1,1\right)$ to part of the circle passing through $1$, $-1$ and $i/2$, with $1$ and $-1$ being its end points. Likewise, $f(1)=1$, $f(-1)=-1$, $f(i)=3i$, for which $f$ maps the upper-half circle to part of the circle passing through $1$, $-1$ and $3i$. Now you can figure out the boundary. One more step is to determine if the image is the interior points of your boundary or the exterior points of your boundary. As such, check $f(i/2)=4i/3$. Therefore, the image must be the interior points of your boundary. Apr 15, 2018 at 10:56
• Thank you! So if I got it right, the image will be a rectangle $(-1,1)\times(i/2,3i)$. What is actually happens is that upper-half circle transforms into this rectangle right? Apr 15, 2018 at 11:20

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.