Prove that if $|z+w|=|z-w|$ then $z\overline{w}$ is purely imaginary.

Question

Prove that if $|z+w|=|z-w|$ then $z\overline{w}$ is purely imaginary.

To start off, I said let $z=a+bi$ and let $w=p+qi$. Not sure where to go from here after subbing in those for $z$ and $w$.

Answer 1

You don't even need to explicit $z$ and $w$. You just have to show that $$\overline{z\,\overline w}=-z\,\overline w.$$

You can use that $|a|=|b|\iff a\,\overline a=b\,\overline b$.

Answer 2

$$|z+\omega|^2=|z-\omega|^2 \iff$$ $$(z+\omega)(\bar{z}+\bar{\omega})= (z-\omega)(\bar{z}-\bar{\omega})\iff$$ $$z\bar{z}+{\omega}\bar{z}+z\bar{\omega}+\bar{\omega}\omega= z\bar{z}-\omega\bar{z}-z\bar{\omega}+\bar{\omega}\omega\iff$$ $$2\omega\bar{z}=-2\bar{\omega}z\iff$$ $$\omega\bar{z}=-\overline{\omega\bar{z}}$$

Answer 3

$z + w = (a + p) + (b + q)i$

$|z + w|^2 = (a + p)^2 + (b + q)^2$

$z - w = (a - p) + (b - q)i$

$|z - w|^2 = (a - p)^2 + (b - q)^2$

$a^2 + 2ap + p^2 + b^2 + 2bq + q^2 = a^2 - 2ap + p^2 + b^2 - 2bq + q^2$

$4ap + 4bq = 0$

$ap + bq = 0$ [1]

$z\bar{w} = (a + bi)(p - qi) = (ap + bq) + i(bp - aq) = (bp - aq)i$, which is purely imaginary (using [1])