how to find the image of the infinite vertical strip

Question

how to find the image of the infinite vertical strip defined by $-2 \leq x \leq -1 , \infty

Answer 1

Note that Möbius maps take circles/lines to circles/lines, and they're continuous on the extended complex plane. So you just need to find the image of the two lines $x=-2$ and $x=-1$ (these images will be circles or lines, and they will bound some region of the extended complex plane). Then you need to determine which one of the two regions of the extended complex plane is the image of the strip, and which is the image of the complement of the strip.

Answer 2

If you don't know the properties of Mobius transformations, you can solve the problem using the polar representation of complex numbers (and the equation of a circle in polar coordinates that you can see here).

A complex number $p=-1+iy$ on the line $x=-1$ can be represented in polar form as: $$ p=\frac{1}{\cos(\pi-\theta)}e^{i\theta}\quad \mbox{with}\quad \frac{\pi}{2}\le\theta \le \frac{3\pi}{2} $$ so we have: $$ \frac{1}{p}=\left( \frac{-1}{\cos(\pi-\theta)}e^{i\theta}\right)^{-1}=\cos(\pi- \theta) e^{-i\theta} $$ So the modulus of $\frac{1}{p}$ is $$ \rho=\cos(\pi-\theta) $$ and this is the equation, in polar coordinates, of a circle with center at the point $(-\frac{1}{2},0)$ and radius $r=\frac{1}{2}$.

In the same way, starting from a complex number $q=-2+iy$ on the line $x=-2$, you can find that this line is transformed by $q \rightarrow \frac{1}{q}$ to the circle with center at $(-\frac{1}{4},0)$ and radius $r=\frac{1}{4}$.

So the vertical strip between the two lines is transformed in the ''lunula'' between the two circle $c$ and $d$ in the figure.