# inversion of the circle $t \mapsto (3 + is) + e^{it}$ around the unit circle.

We know that inversion interchanges lines and circles, but it's very hard to

The inversion map about the unit circle is just $\displaystyle z \mapsto \frac{1}{\overline{z}}$. As a Möbius transformation maps circles to circles, let's try a circle with radius $1$ centered on the line $\mathrm{Re}(z)=3$:

$$\big\{ z(t)= (3 + is) + e^{it} : 0 \leq t \leq 2 \pi \big\}$$

The image equation is easy write down, but to recognize but hard to identify as a circle:

$$\frac{1}{(3-is)+e^{-it}}$$

What is the radius? Or the center? If $s = 0$ I thought I could make some headway. It is just as bad:

$$\frac{1}{3+e^{-it}} = \frac{3 + e^{it}}{10 + 6\cos t}$$

Assuming the image is a circle, we can just notice that $2 \mapsto 1/2$ and $4 \mapsto 1/4$ so the new diameter is $\frac{1}{2}-\frac{1}{4} = \frac{1}{4}$ with the center at $( \frac{3}{8} ,0)$.

In the case of the line, a circle of radius $r = \infty$, it is possible to show by reparamerization that $z = 3 + it$ maps to the circle of radius $\frac{1}{3}$ and center $(0, \frac{1}{6})$. I don't know how to do that in this case.

The given circle can be written as $|z-\alpha|=1$, where $\alpha=3+is$. The image points $w$ satisfy $$\Bigl|\frac1w-\alpha\Bigr|=1$$ which can be written $$|1-\alpha w|=|w|\ .$$ Now square both sides and write in terms of conjugates: $$\def\c#1{\overline{#1}}1-\alpha w-\c\alpha\c w+\alpha\c\alpha w\c w=w\c w\ .$$ Collecting terms and dividing by $\alpha\c\alpha-1$ (which is not zero) gives $$w\c w-\frac\alpha{\alpha\c\alpha-1}w-\frac{\c\alpha}{\alpha\c\alpha-1}\c w+\frac1{\alpha\c\alpha-1}=0$$ and hence $$\Bigl(w-\frac{\c\alpha}{\alpha\c\alpha-1}\Bigr) \Bigl(\c w-\frac\alpha{\alpha\c\alpha-1}\Bigr)=\langle\hbox{something}\rangle\ .$$ See if you can supply the missing RHS and finish the problem.
To complement David's nice algebraic derivation, notice to invert $|z - a| = 1$ about the unit circle, we can rotate the plane by phase $a/|a|$ so the center lies on the real axis.
• $$r = \frac{1}{2} \left( \frac{1}{|a|-1} - \frac{1}{|a|+1}\right) = \frac{1}{|a|^2 - 1}$$
• $$z_0 = \frac{1}{2} \left( \frac{1}{|a|-1} + \frac{1}{|a|+1}\right) = \frac{|a|}{|a|^2 - 1}$$
For a general circle we will have $a$ instead of $|a|$ in the numerator.