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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)

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    Yes, that's what I mean.2012-10-28

1 Answers 1

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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?