Image of a complex disc by this function? [closed]

Question

I am working with the function $f(z) = \frac{z+1}{z-1}$, for a complex variable $z$. I understood that for $z$ in the unit disc, i.e $\lvert z\rvert \le 1$, $\mathrm{Re}(f(z)) \le 0$.

What if $z$ is in a disc, say $D = \left\{ z \in \mathbb C: \, \lvert z\rvert \le a \right\}$ for a given constant $a \in \mathbb R_{+}$?

Note that the function is its own inverse, if it matters …. I'm not very good with analysis and I have trouble visualizing functions from $\mathbb C$ to $\mathbb C$, since there are 4 dimensions ….

Answer 1

This question might have been better suited to math.SE but since it's here I may as well post an outline of an answer. Your function $f$ is an example of a Moebius transformation, and these are known to take circles to circles provided that you allow straight lines as circles passing through "the point at infinity". So for $0<a<1$ the image of your closed disc is another closed disc, living in the left half-plane; for $a>1$ you get the complement of some open disc that lives in the right half-plane.

Answer 2

You don't need to think in 4-D, just put two complex planes, one the domain the other the codomain, next to each other. To get an idea of the action of the map you can begin by plotting points in the domain and their images in the codomain. Next you can put in all real numbers and see what is done to that, or the imaginary numbers (they get mapped onto the unit circle. If you want to know what happens to your other discs parametrize the boundary circle, say $z=a\exp(it)$ with $0\le t\le2\pi$, and calculate the image curve; you'll find it's another circle.

As you remarked the map is its own inverse, so the image of the circle of radius $a$ around the origin is also equal to the set of $z$ for which $(z+1)/(z-1)$ has modulus $a$, or $\lvert z+1\rvert=a\lvert z-1\rvert$. Now just plug in $z=x+iy$ and square both sides to get rid of the square root to get $(x+1)^2+y^2=a^2((x-1)^2+y^2)$, from there you can create an equation for the circle that will show centre and radius. (In the case $a=1$ the squares disappear and you end up with the imaginary axis, a circle with radius $\infty$.)

If you've developed some skill with the Moebius transforms mentioned in the other answer you'll know that circles and lines get mapped to circles and lines and that all kind of symmetry is preserved. Then you'd know that the image must be a circle because it intersects the real axis twice, it intersects it perpendicularly beceause the original circle does, hence its centre is on the real axis and necessarily halfay between $(a+1)/(a-1)$ and $(-a+1)/(-a-1)$.