Complex Numbers are Perpendicular iff Dot Product is Zero

Theorem

Let $z_1$ and $z_2$ be complex numbers in vector form such that $z_1 \ne 0$ and $z_2 \ne 0$.

Then $z_1$ and $z_2$ are perpendicular if and only if:

$z_1 \circ z_2 = 0$

where $z_1 \circ z_2$ denotes the complex dot product of $z_1$ with $z_2$.

Proof

By definition of complex dot product:

$z_1 \circ z_2 = \cmod {z_1} \, \cmod {z_2} \cos \theta$
$\cmod {z_1}$ denotes the complex modulus of $z_1$
$\theta$ denotes the angle from $z_1$ to $z_2$, measured in the positive direction.

Necessary Condition

Let $z_1$ and $z_2$ be perpendicular.

Then either $\theta = 90^\circ$ or $\theta = 270^\circ$.

Either way, from Cosine of Right Angle or Cosine of Three Right Angles:

$\cos \theta = 0$

and so:

$\cmod {z_1} \, \cmod {z_2} \cos \theta = 0$

Hence by definition:

$z_1 \circ z_2 = 0$

$\Box$

Sufficient Condition

Let $z_1 \circ z_2 = 0$.

Then by definition:

$\cmod {z_1} \, \cmod {z_2} \cos \theta = 0$

As neither $z_1 = 0$ or $z_2 = 0$ it follows that $\cos \theta = 0$.

From Cosine of Half-Integer Multiple of Pi it follows that either:

$\theta = 90^\circ$
$\theta = 270^\circ$

or:

$\theta$ is either of the above plus a full circle.

That is, $z_1$ and $z_2$ are perpendicular.

$\blacksquare$