If $z$ and $w$ are complex numbers such that $|z+w| = |z-w|$
Prove that $\arg z - \arg w = \pm \ \pi/2$
Can someone please help me?
If $z$ and $w$ are complex numbers such that $|z+w| = |z-w|$
Prove that $\arg z - \arg w = \pm \ \pi/2$
Can someone please help me?
$|z+w|^2 = |z|^2+2 \operatorname{re} \overline{z} w + |w|^2$.
$|z-w|^2 = |z|^2-2 \operatorname{re} \overline{z} w + |w|^2$.
If the two are equal, we must have $\operatorname{re} \overline{z} w = 0$.
What does that say about $\operatorname{arg} \overline{z} w$?