# $(z+\bar z)|w|^2 - (z-\bar z)|w|i - 2(z + \bar z) = 0$

I have the following equation:

$$(z+\bar z)|w|^2 - (z-\bar z)|w|i - 2(z + \bar z) = 0.$$

I need to show that $z$ can't be an imaginary number.

then that the image of $z$ passes from (0,0)

-
by $z^*$ do you mean the complex conjugate? then it is better to use $\bar z$ –  Dennis Gulko Oct 28 '12 at 14:52
Yes, that's what I mean. –  Chris Oct 28 '12 at 14:53

By rearranging you equation, we get $$(z+\bar z)|w|^2 - 2(z + \bar z) = (z-\bar z)|w|i$$ Do you know that $z+\bar z=2\Re(z)$ and $z-\bar z=2i\Im(z)$ - if you never seen this before, prove it by setting $z=x+iy$ and computing directly. So you have: $$2(|w|^2-2)\Re(z)=-2|w|\Im(z)$$ What happens if $z$ is imaginary?