Möbius transformation image

Question

Let $f(z)=\frac{az+b}{z+d}$, when $d\in\mathbb{R}$, $d\not=0$ $a,b\in\mathbb{C}$ and $f$ is not constant. I want to find the image of the real and imaginary axes under $f$. I've found that the image of the real axis is $\{z:z=\frac{a+b}{1+d}+\frac{ad-b}{d(1+d)}u,u\in\mathbb{R}\}$. The pole of f is not on the imaginary axis therefore his image is a circle. From the angle preservation principle the diameter of the circle is on the image of the real axis. I would like a hint.

Answer 1

Your Möbius transformation goes from the $z$-plane to the $w$-plane, where

$$w = \frac{az+b}{z+d}$$

You need to rearrange this equation to make $z$ the subject. Let's say you have $$z=\frac{pw+q}{rw+s}$$

An equation for the imaginary axis in the $z$-plane is $|z-1|=|z+1|$. (The imaginary axis is the perpendicular bisector of the chord joining $-1$ and $1$.) This equation then becomes $$\left|\frac{pw+q}{rw+s}-1\right| = \left|\frac{pw+q}{rw+s}+1\right|$$ The solution set of this equation is something in the $w$-plane. It is the image of the imaginary axis. Make the substitution $w=u+\mathrm{i}v$, apply the definition of modulus, simplify, complete the square on $u$ and $v$ and then read off the answer. See my example below:

Example

Consider the Mobius transformation from the $z$-plane to the $w$-plane where

$$w=\frac{z+1}{z+2} \implies z=\frac{1-2w}{w-1}$$

Now we substitute this expression for $z$ into $|z-1|=|z+1|$ and then substitute $w=u+\mathrm{i}v$ $$\begin{eqnarray*} |z-1| &=& |z+1| \\ \\ \left|\frac{1-2w}{w-1}-1\right| &=& \left|\frac{1-2w}{w-1}+1\right| \\ \\ \left|\frac{2-3w}{w-1}\right| &=& \left|\frac{-w}{w-1}\right| \\ \\ \frac{|2-3w|}{|w-1|} &=& \frac{|-w|}{|w-1|} \\ \\ |3w-2| &=& |w| \\ \\ |(3u-2)+3\mathrm{i}v| &=& |u+\mathrm{i}v| \\ \\ (3u-2)^2 + 9v^2 &=& u^2 + v^2 \\ \\ 9u^2 - 12u + 4 + 9v^2 &=& u^2 + v^2 \\ \\ 8u^2 - 12u +8v^2 + 4 &=& 0 \\ \\ 8\left(u-\frac{3}{4}\right)^2+8v^2 &=& \frac{1}{2} \\ \\ \left(u-\frac{3}{4}\right)^2+v^2 &=& \frac{1}{16} \end{eqnarray*}$$

Finally, we see that the image of the imaginary axis is a circle, centre $w=\frac{3}{4}$ with radius $\frac{1}{4}$.