1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $w$?

$\endgroup$
3
  • $\begingroup$ Can't understand how you got to the last equality. $\endgroup$
    – UserX
    Sep 20 '14 at 15:45
  • $\begingroup$ If $c$ is real, then the imaginary part of $cz$ is $c$ times the imaginary part of $z$. @UserX $\endgroup$ Sep 20 '14 at 15:50
  • $\begingroup$ @UserX: Edited to try to make things clearer (and somehow made it even more terse than before.) $\endgroup$ Sep 20 '14 at 15:57
1
$\begingroup$

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: $$\omega-\bar\omega=i$$

$\endgroup$
1
  • $\begingroup$ I did this and left it there... should be $a^2+b^2=2b$ up there and $w=\frac{a}{2b}-\frac12 i$ $\endgroup$
    – UserX
    Sep 20 '14 at 15:40

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.