## Monday Math 147

Given two non-zero complex numbers z1 and z2 such that $\left|z_1+z_2\right|=\left|z_1-z_2\right|$, show that the arguments of z1 and z2 differ by π/2.

Recall that for any complex number z, $\left|z\right|^2=z\bar{z}$, thus
$\begin{array}{rcl}\left|z_1+z_2\right|^2&=&\left|z_1-z_2\right|^2\\\left(z_1+z_2\right)\left(\bar{z_1}+\bar{z_2}\right)&=&\left(z_1-z_2\right)\left(\bar{z_1}-\bar{z_2}\right)\\z_1\bar{z_1}+z_1\bar{z_2}+\bar{z_1}z_2+z_2\bar{z_2}&=&z_1\bar{z_1}-z_1\bar{z_2}-\bar{z_1}z_2+z_2\bar{z_2}\\z_1\bar{z_2}+\bar{z_1}z_2&=&-z_1\bar{z_2}-\bar{z_1}z_2\\z_1\bar{z_2}+\bar{z_1}z_2&=&{0}\end{array}$

Now, recall that for any complex number z, $z+\bar{z}=2\Re(z)$, and since $\overline{z_1\bar{z_2}}=\bar{z_1}z_2$, we see that
$z_1\bar{z_2}+\bar{z_1}z_2=2\Re\left(z_1\bar{z_2}\right)$, and so
$\Re\left(z_1\bar{z_2}\right)=0$,
and we proved last week that z1 and z2 have arguments differing by π/2 (and thus give perpendicular vectors) if and only if $\Re\left(z_1\bar{z_2}\right)=0$.

Consider the geometric interpretation. Given a parallelogram with sides given by the vectors corresponding to z1 and z2, then z1+z2 and z1z2 correspond to the diagonals; thus, the above is equivalent to stating that if the diagonals of a parallelogram are of equal length, then the parallelogram is a rectangle.