# The image of the curve with equation $z\bar z = 2\operatorname{Im} z$ under the map $w=1/z$

Let $z \in \mathbb{C}$ satisfy $z \overline{z} = 2\Im (z)$

Let $w=\frac{1}{z}$

Find the equation describing the curve $w$ forms on the complex plane and the $z$ that has the minimun distance from that curve.

I found that $z$ forms a circle with center $C(0,1)$ and radius $r=1$ and I can't find how to connect these it to $w$. I would prefer a full solution if you have the time, but a hint is fine too.

## 2 Answers

Hint: Since $z\overline{z}$ is real, we may write $\frac{1}{2}=\frac{\Im(z)}{z\overline{z}}=\Im(\frac{z}{z\overline{z}})=\Im\left(\frac{1}{\overline{z}}\right).$ What is this in terms of $w$?

• Can't understand how you got to the last equality. Sep 20 '14 at 15:45
• If $c$ is real, then the imaginary part of $cz$ is $c$ times the imaginary part of $z$. @UserX Sep 20 '14 at 15:50
• @UserX: Edited to try to make things clearer (and somehow made it even more terse than before.) Sep 20 '14 at 15:57

Let $z:=a+ib$ $$z\bar z=2\Im(z)\equiv(a+ib)(a-ib)=2b\implies a^2+b^2=2b$$ Now $$\omega=\frac1z=\frac1{a+ib}=\frac{a-ib}{a^2+b^2}=\frac a{2b}-\frac12i$$ So: $$\omega-\bar\omega=i$$

• I did this and left it there... should be $a^2+b^2=2b$ up there and $w=\frac{a}{2b}-\frac12 i$ Sep 20 '14 at 15:40