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 $|z-a| = |z-b|$?

  • $\begingroup$ Your set ls always a circle (possibly of radius $0$). $\endgroup$ – Kabo Murphy May 16 at 9:25
  • $\begingroup$ Strictly speaking, inversion is $z \to 1/\bar z$ in order that $O,M,M'$ are aligned. See paragraph 4 in math.stackexchange.com/q/2626356 $\endgroup$ – Jean Marie May 19 at 19:55

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$$(\forall t\in\mathbb R):\left(\frac t{2t^2-2t+1}-\frac12\right)^2+\left(\frac{t-1}{2t^2-2t+1}+\frac12\right)^2=1.$$

  • $\begingroup$ 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... $\endgroup$ – PhysicsMathsLove May 16 at 9:34
  • $\begingroup$ 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. $\endgroup$ – José Carlos Santos May 16 at 9:44
  • $\begingroup$ ok it just that this was in an exam and I have a complex analysis exam today $\endgroup$ – PhysicsMathsLove May 16 at 9:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.