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How to find the inverse point of the point z=a with respect to the circle $|z-c|=r$ (where c is the origin and r the radius) ?

$c+\frac{r^{2}}{a-c}$

this is the answer given in the book...how do we get it

Gerry Myerson
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amit
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  • I won't do the calculations, but here is the principle. Find a Möbius transformation $\sigma \colon z \mapsto (az + b)/(cz + d)$ that transforms the given circle into the real line. Apply $\sigma$ to $a$, then take the reflection of $\sigma(a)$ with respect to the real axis (its complex conjugate), and finally apply $\sigma^{-1}$ to the result. – Dave Sep 08 '14 at 06:24
  • ohk...i'll try with this – amit Sep 08 '14 at 06:35
  • Shouldn't that be $\overline{a-c}$, not $a-c$ (ie the complex conjugate)? – almagest Sep 08 '14 at 08:16
  • An easier way is to start by figuring out the inverse point of $z$ wrt the unit circle centred at 0 (rather than $c$). – almagest Sep 08 '14 at 08:17
  • i dint get it..can anyone solve it? – amit Sep 08 '14 at 12:18

2 Answers2

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If $z_0,p,q$ are collinear points, where $z_0$ is center of circle with radius $k$, $q$ lies outside the circle, and $p$ is inverse point of $q$ which lies inside the circle, then $$p=\frac{z_0+k^2(q-z_0)}{|q-z_0|^2}$$

$$P =\frac{z_0+k^2(q-z_0)}{(q-z_0)(q-Z_0)}$$

$$P=\frac{z_0+k^2}{q-z_0}$$

daOnlyBG
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soni
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Start by noting that the inverse point of $a$ wrt the unit circle centre the origin is $1/\,\overline{a}$. Hence the inverse point of $a$ a circle radius $r$ centre the origin is $\frac{r^2}{\overline{a}}$. Now we have to translate that to a circle centre $c$. Relative to $c$, the point $a$ is $a-c$, so we want $c+\frac{r^2}{\overline{a-c}}$.

almagest
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  • Why the coniugation? Isn't it just $1/a$? – Lolman Sep 08 '14 at 12:58
  • Think about it. What happens if $a=i$ for example? More generally if $\arg z=\theta$, then $\arg\frac{1}{z}=-\theta$ – almagest Sep 08 '14 at 13:02
  • exactly: $$arg\frac{1}{z}=arg\frac{\bar{z}}{z*\bar{z}}=arg(\bar{z})=-\theta$$ Is it not? – Lolman Sep 08 '14 at 13:07
  • Yes. Exactly. Both $1/z$ and $\overline{z}$ have the same arg. So $z$ and $1/\overline{z}$ have the same arg. – almagest Sep 08 '14 at 13:10
  • I think when he asked for an inverse he meant the multiplicative inverse. Looking at the answer his book gave is more than enough. – Lolman Sep 08 '14 at 13:15
  • @Lolman. I doubt it. He is asking for the inverse wrt a circle whose centre is not the origin. A multiplicative inverse would not make much sense. – almagest Sep 08 '14 at 13:19