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I'm trying to prove that the inversion mapping $f(z) = \frac{1}{z}$ sends circles or lines to circles or lines.

Apparently the set $$\{z \in \mathbb{C}: |z-a|^2 = r^2 \}$$ describes either a circle or a line.

How does this set includes circle and lines?

Circles are defined as $|z - a| = r$ and lines are defined as $|z-a| = |z-b|$?

José Carlos Santos
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PhysicsMathsLove
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1 Answers1

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That set describes a circle, not a line.

The function $f$ maps circles which pass through the origin into lines and all other circles into circles. It als maps line passing through the origin into lines and all other lines into circlse.

Consider, for instance, the line $\{t+(1-t)i\,|\,t\in\mathbb R\}$. For each $t\in\mathbb R$, you have$$\frac1{t+(1-t)i}=\frac{t+(t-1)i}{t^2+(1-t)^2}=\frac t{2t^2-2t+1}+\frac{t-1}{2t^2-2t+1}i$$and$$(\forall t\in\mathbb R):\left(\frac t{2t^2-2t+1}-\frac12\right)^2+\left(\frac{t-1}{2t^2-2t+1}+\frac12\right)^2=1.$$

José Carlos Santos
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  • However, my lecturer says that the complex representation of a line is $l = { z \in \mathbb{C}: \overline{a} z + a\overline{z} = b, a \in \mathbb{C}, b \in \mathbb{R}} $. Do you not get this if you write $|z-a|^2 = r^2 \Rightarrow (z-a)\overline{(z-a)} = r^2$ and expand? Edit: the latex isn't going my way (can't format set brackets) but hopefully you can see what I mean... – PhysicsMathsLove May 16 '19 at 09:34
  • Do you expect me to reply saying the opposite of what I wrote in my answer? I wrote that ${z\in\mathbb C,|,\lvert z-a\rvert^2=r^2}$ is a circle, not a line, which is something that you can check for yourself. – José Carlos Santos May 16 '19 at 09:44
  • ok it just that this was in an exam and I have a complex analysis exam today – PhysicsMathsLove May 16 '19 at 09:45