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I'm studying for my exam and one of the questions I am stuck on is:

Show that under inversion in the unit circle a circle with centre C and radius $S$ inverts into a circle with centre $\frac{C}{C\overline{C} - s^2}$ and radius $\frac{s}{C\overline{C} - s^2}$.

And I know: $$Z\overline{Z}-Z\overline{C}-\overline{Z}C+C\overline{C}-s^2=0$$

Can someone please explain how to do this question?

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  • $\begingroup$ What is $Z$ mapped to under inversion in the unit circle? $\endgroup$ – Daniel Fischer Oct 27 '13 at 9:55
  • $\begingroup$ I don't know?? :( $\endgroup$ – Jay C Oct 27 '13 at 10:15
  • $\begingroup$ What is, geometrically, the inversion in the unit circle? $\endgroup$ – Daniel Fischer Oct 27 '13 at 10:17
  • $\begingroup$ is that i...? or what is it? $\endgroup$ – Jay C Oct 27 '13 at 10:38
  • $\begingroup$ Inversion in a circle is a map. Can you describe that map geometrically? $\endgroup$ – Daniel Fischer Oct 27 '13 at 10:40
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Inversion is

$$z\mapsto \frac1{\bar z}$$

Your equation for the circle is basically the squared form of

$$ \lvert z - C\rvert = s $$

Now combine them:

\begin{align*} \left\lvert\frac1{\bar z}-C\right\rvert&=s \\ \left(\frac1{\bar z}-C\right)\left(\frac1{z}-\bar C\right)&=s^2 \\ \frac1{z\bar z}-\frac{C}{z}-\frac{\bar C}{\bar z}+C\bar C-s^2&=0 \\ 1-C\bar z -\bar Cz+z\bar z\left(C\bar C-s^2\right) &= 0 \\ z\bar z - \frac{C}{C\bar C-s^2}\bar z - \overline{\left(\frac{C}{C\bar C-s^2}\right)}z + \frac{1}{C\bar C-s^2} &= 0 \\ \left(z-\frac{C}{C\bar C-s^2}\right) \left(\bar z-\frac{\bar C}{C\bar C-s^2}\right) -\frac{C\bar C}{\left(C\bar C-s^2\right)^2} +\frac{C\bar C-s^2}{\left(C\bar C-s^2\right)^2} &= 0 \\ \left(z-\frac{C}{C\bar C-s^2}\right) \left(\bar z-\frac{\bar C}{C\bar C-s^2}\right) -\left(\frac{s}{C\bar C-s^2}\right)^2 &= 0 \\ \left(z-\frac{C}{C\bar C-s^2}\right) \left(\bar z-\frac{\bar C}{C\bar C-s^2}\right) &= \left(\frac{s}{C\bar C-s^2}\right)^2 \\ \left\lvert z-\frac{C}{C\bar C-s^2}\right\rvert &= \frac{s}{C\bar C-s^2} \end{align*}

From this you can read

\begin{align*} C' &= \frac{C}{C\bar C-s^2} & s' &= \frac{s}{C\bar C-s^2} \end{align*}

as expected.

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  • $\begingroup$ MvG, I feel like you always help me out in this class. I wish you could tutor me ;-; Thanks so much!!! $\endgroup$ – Jay C Oct 27 '13 at 23:50
  • $\begingroup$ MvG, I was wondering if I have a circle C and radius r and centre o = (1,1), how does that change things? I have to find the radius of the circle C' (in terms of r) which is the image of C after inversion in the unit circle about the origin. $\endgroup$ – Jay C Nov 1 '13 at 7:51
  • $\begingroup$ @JayC: I don't understand your question. Simply apply the above formulas, I don't see why anything should change. Except for the fact that in the above formulas, $C$ is the center and $s$ the radius, instead of center $o$ and radius $r$. And you'd have to write $o=1+i$ instead of $o=(1,1)$, obviously. In any case, I don't feel much inclined to answer new questions unless you start accepting those answers which did help you, as you claim they did. $\endgroup$ – MvG Nov 1 '13 at 12:54

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