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.

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