Find the positions relative to a set.

Question

Find positions of $\dfrac{1}z$, $\bar{z}$ and $-z$ on a set $|z|=2^2$?

The position of $\dfrac{1}z$ will be $x^2+y^2=\dfrac12$ and $\bar{z}$ will be $x^2+y^2=2$ (which have no effect) and $-z=|-z|=|z|$ (which also have no effect).

Can someone please confirm if I'm right.

|z|=sqrt(x^2+y^2)=2^2, don't know where you get 16? – Sam123 Aug 23 '13 at 11:14 — Sam123, Aug 23 '13 at 11:14

Rasmus · Answer 1 · 2013-08-23T11:31:16.473

1

Well, the position of $-z$ will be opposite to the one of $z$ on the circle of radius $4$ around $0$. The position of $\bar z$ is the one of $z$ reflected at the $x$-axis. Finally, since $$\frac1z=\frac{\bar z}{\lvert z\rvert^2}=\frac{\bar z}{\lvert\bar z\rvert^2},$$ we get the position of $\frac1z$ by starting from $z$, reflecting at the $x$-axis, and then inverting the distance to the origin.

edited Aug 23 '13 at 11:31

answered Aug 23 '13 at 11:25

Rasmus

18,404

Let $z=r e^{i\phi}$, then $$\frac{1}{z} = r^{-1} e^{-i\phi}$$ so actually $$\frac{1}{z} = \frac{\bar{z}}{|z|^2}$$. Or in other words $|z|^2 = z\bar{z} \neq z^2$ – AlexR Aug 23 '13 at 11:27
@AlexR corrected, thanks. – Rasmus Aug 23 '13 at 11:31
Good. Now it's correct :-) - removed my -1. – AlexR Aug 23 '13 at 11:32

score 0 · Answer 2 · answered Aug 23 '13 at 11:25

You have $A=\partial B_4(0)$, so $A = -A = \overline{A}$, $$A^{-1} = \{ (re^{i\phi})^{-1} : r = 4, \phi \in [0, 2\pi) \} = \{ r^{-1} e^{-i\phi} : r = 4, \phi \in [0, 2\pi) \} = \{ w e^{i\theta} : w = \frac{1}{4}, \theta \in (-2\pi, 0] \} = \partial B_{\frac{1}{4}}(0)$$ So you are right.

Find the positions relative to a set.

2 Answers2