Möbius transformation, determine the image area for the w-plane

Question

Can someone help me with this question.

Let

$D\{ z\in \mathbb{C}: Re(z)<0,|z-1|<2\}$

and look at the Möbius transformation

$f(z)=\frac{z-i}{z+1}$.

Determine the image area

$D'=\{f(z):z\in D\}$.

I have started by drawing the circle $|z-1|=2$ in the z-plane, then determine the area of the circle that fulfills the conditions $Re(z)<0, |z-1|<2$. See the figure (excuse my bad drawing).

And then I was thinking that I could choose 3 points that has this area on the left: $z_{1}=-i, z_{2}=0, z_{3}=i$ and then calculate the corresponding $w_{1}, w_{2}, w_{3}$ by using the formula above $f(z)=w$.

Then I get $w_{1}=-i+1$, $w_{2}=-i$ and $w_{3}=0$, how can I draw the image area for the w-plane from this? Do I have to do something else?

Answer 1

In your set-up, your region of interest is bounded by a circle and a line. The circle has real equation: $$ (x-1)^2+y^2=4 $$ and the line has real equation: $$ y=0. $$

Mobius transformations (and circle inversions, in more generality) take generalized circles to generalized circles. The phrase generalized circles is just a name for circles and lines.

So, in your example, you've got to find the generalized circles that your objects transform into.

• For $(x-1)^2+y^2=4$, we can take the points $z=-1$, $3$, and $1+2i$. By applying the transformation to each of these three points, we get

  1. $f(-1)=\frac{-1-i}{-1+1}=\infty$.

  2. $f(3)=\frac{3-i}{3+1}=\frac{3}{4}-\frac{1}{4}i$

  3. $f(1+2i)=\frac{1+2i-i}{1+2i+1}=\frac{1+i}{2+2i}=\frac{1}{2}$.

Since one of the points on the boundary of the circle is taken to infinity, this circle maps to a line. Moreover, the line passes through $\frac{1}{2}$ and $\frac{3}{4}-\frac{1}{4}i$. This means that you're looking at the line $y=-x+\frac{1}{2}$.

• For $y=0$, three nice points are $z=0$, $z=i$, and $z=\infty$. By applying the transformation to each of these three points, we get

  1. $f(0)=\frac{0-i}{0+1}=-i$.

  2. $f(\infty)=\frac{\infty-i}{\infty+1}=1$.

  3. $f(i)=\frac{i-i}{i-1}=0$.

Therefore, this line becomes the circle passing through $0$, $1$, and $-i$. The center of the circle must be on the bisector of the segment between $0$ and $1$, so its real part is $\frac{1}{2}$. Similarly, the center of the circle must be on the bisector of the segment between $0$ and $-i$, so its imaginary part is $-\frac{1}{2}i$. Hence, the image of the line is the circle $\left(x-\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2=\frac{1}{2}$. (The $\frac{1}{2}$ on the RHS comes from recalling that $0$ is on the circle.)

So, the region of interest for you is one of the four regions determined by this line and circle. Since the boundary of your region includes the point $-1$, which is mapped to $\infty$, this should be an unbounded region. So, the area should be infinite.