## Monday Math 146

Consider two non-zero complex numbers z1 and z2. If the arguments of these two numbers differ by π/2 (the complex numbers, when treated as vectors in the complex plane, define perpendicular vectors), what does this say about the value of $z_1\bar{z_2}$?

First, we note that $\arg\bar{z_2}=-\arg{z_2}$, and that the argument of the product of two complex numbers is the sum of the arguments of the numbers. Thus,
$\arg\left(z_1\bar{z_2}\right)=\arg{z_1}+\arg\bar{z_2}=\arg{z_1}-\arg{z_2}$
Now, we said that the two arguments differ by π/2; thus
$\arg\left(z_1\bar{z_2}\right)=\arg{z_1}-\arg{z_2}=\pm\frac{\pi}2$
but that is the argument of a purely imaginary number; so
$\Re\left(z_1\bar{z_2}\right)=0$.
We see that this argument is reversible; if $\Re\left(z_1\bar{z_2}\right)=0$, then
$\arg\left(z_1\bar{z_2}\right)=\arg{z_1}-\arg{z_2}=\pm\frac{\pi}2$,
and the vectors representing two nonzero complex numbers z1 and z2 are perpendicular if and only if $\Re\left(z_1\bar{z_2}\right)=0$.

Note that if we give $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$, then
$\begin{array}{rcl}\Re\left(z_1\bar{z_2}\right)&=&\Re\left[\left(x_1+iy_1\right)\left(x_2-iy_2\right)\right]\\&=&\Re\left[x_1x_2-ix_1y_2+iy_1x_2+y_1y_2\right]\\&=&\Re\left[x_1x_2+y_1y_2+i\left(y_1x_2-x_1y_2\right)\right]\\\Re\left(z_1\bar{z_2}\right)&=&x_1x_2+y_1y_2\end{array}$.
Note that expressed as conventional vectors $\vec{v}_1=(x_1,y_1)$ and $\vec{v}_2=(x_2,y_2)$, then $\vec{v}_1\cdot\vec{v}_2=x_1x_2+y_1y_2$, so
$\Re\left(z_1\bar{z_2}\right)=0$ is equivalent to $\vec{v}_1\cdot\vec{v}_2=0$, confirming perpendicularity of the vectors.