Image of a circle under conformal map $1/z$

Question

The image of a circle under conformal map $1/z$ should be a circle, but how to prove it (or how to find the relationship between the two circles)?

$z = x + iy = d + a\exp(i\theta)$, where $a$ is the radius of the circle and $d$ is a real number, represents a circle with radius $a$ displaced along the real axis.

Now $w = u + iv = 1/z$; how to find the center position and radius of the transformed circle?

I just write $w = 1/(d + a\exp(i\theta))$ and after some algebra I get $$ u = \frac{d + a\cos(\theta)}{d^2 + a^2 + 2d\cos(\theta)}\quad\text{and}\quad v = -\frac{a\cos(\theta)}{d^2 + a^2 + 2d\cos(\theta)}, $$ but then I don't know how to simplify it and get it to the form that looks like a circle.

Answer 1

The real points $d \pm a$ lie on a diameter of your circle. The image points $1/(d \pm a)$ lie on a diameter of the image circle, so the image circle has center $$ \frac{1}{2}\left[\frac{1}{d - a} + \frac{1}{d + a}\right] = \frac{d}{d^{2} - a^{2}} $$ and radius $$ \frac{1}{2}\left|\frac{1}{d - a} - \frac{1}{d + a}\right| = \frac{a}{|d^{2} - a^{2}|}. $$

Alternatively, starting from $$ (x - d)^{2} + y^{2} = a^{2} $$ and substituting $$ x = \frac{u}{u^{2} + v^{2}},\qquad y = -\frac{v}{u^{2} + v^{2}} $$ works.