Image of Vertical Strip under Inversion

This is a slight generalisation of Example 3 in Churchill’s Complex Variables and Applications.

Consider the infinite vertical strip $c_1<x<c_2$ , under the transformation $w=1/z$ , where $c_1, c_2$ are of the same sign.

When $x=c_1$ , note that by arguments similar to earlier analysis, we have $x=\frac{u}{u^2+v^2}$ .

$u^2+v^2=\frac{1}{c_1}u$ , upon completing the square, becomes

$(u-\frac{1}{2c_1})^2+v^2=(\frac{1}{2c_1})^2$ — Circle 1

Similarly the line $x=c_2$ is transformed into:

$(u-\frac{1}{2c_2})^2+v^2=(\frac{1}{2c_2})^2$ — Circle 2

Note that as $x$ gets larger, the radius of the resultant circle gets smaller.

Thus, the resultant image is the (open) region between the two circles.

The converse holds too, i.e. the region between two circles will be mapped back to the vertical strip by inversion.

Example

Find an analytic isomorphism from the open region between the two circles $|z|=1$ and $|z-\frac{1}{2}|=\frac{1}{2}$ to the vertical strip $0<\text{Re}(z)<1$ .

First, we “shift” the circles to the left by 1 unit via the map $w=z-1$ . Now we are in the situation of our above analysis, thus inversion $w=1/z$ maps the region to the vertical strip $-1<x<-1/2$ . The map $w=-z$ (reflection), followed by $w=z-\frac{1}{2}$ , finally finishing up with a scaling of factor 2 brings us to the vertical strip desired.

Composition of all the above functions leads us to the desired transformation $f(z)=2(\frac{-1}{z-1}-\frac 12)=\frac{-1-z}{z-1}$ to the very nice strip $0<\text{Re}(x)<1$ .