Re: Find solutions of z bar = z^2

Quote:

Originally Posted by

**terrorsquid** Im trying find all solutions for: $\displaystyle \bar{z}=z^2$

Work in polar form.

$\displaystyle \overline{z}=z^2$,

so:

$\displaystyle |\overline{z}|=|z|=|z^2|=|z|^2$

so for non-zeros solutions $\displaystyle |z|=1$, and $\displaystyle z=e^{i\theta}$ for some $\displaystyle \theta \in [0,\2\pi)$

Then:

$\displaystyle \overline{z}=e^{-i\theta}=z^2=e^{2i\theta} $

etc

CB

Re: Find solutions of z bar = z^2

Can someone explain how $\displaystyle |\bar{z}|$ gives you $\displaystyle |z|^2$ ? I can see that $\displaystyle z\bar{z} =|z|^2$ but how does $\displaystyle |z| = |z|^2$ ?

Re: Find solutions of z bar = z^2

Quote:

Originally Posted by

**terrorsquid** Can someone explain how $\displaystyle |\bar{z}|$ gives you $\displaystyle |z|^2$ ? I can see that $\displaystyle z\bar{z} =|z|^2$ but how does $\displaystyle |z| = |z|^2$ ?

You are looking for solutions to:

$\displaystyle \overline{z}=z^2$

now just take absolute values and use $\displaystyle |z^2|=|z|^2.$

CB

Re: Find solutions of z bar = z^2

Oh, right. I think I misread an equation earlier. $\displaystyle \bar{z} = |z|$ so the new equation is:

$\displaystyle |z| = |z|^2$

I read a comment as $\displaystyle |z| = |z|^2$ as an identity separate to my equation - nevermind :S

Thanks.

Out of curiosity how would you derive these solutions if I converted the original equation into:

$\displaystyle r(cos(\theta) - isin(\theta)) = [r(cos(\theta)+isin(\theta))]^2$

can you?

Re: Find solutions of z bar = z^2

Quote:

Originally Posted by

**terrorsquid** So applying this geometrically I would have a solution set of $\displaystyle cos(0+\frac{2\pi k}{3})$ where $\displaystyle k = 0, 1, 2$

Which is the same as the $\displaystyle z = 1$, $\displaystyle z=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$ and $\displaystyle z=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

But where does the z = 0 fit in?

These are the solutions assuming z was NOT 0. Since you have already determined that z= 0 satisfies the original equation, it has **four** solutions.