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 iff:
 * $z_1 \circ z_2 = 0$

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

Proof
By definition of complex dot product:
 * $z_1 \circ z_2 = \left\vert{z_1}\right\vert \, \left\vert{z_2}\right\vert \cos \theta$
 * $\left\vert{z_1}\right\vert$ 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 and Cosine of Three Right Angles:
 * $\cos \theta = 0$

and so:
 * $\left\vert{z_1}\right\vert \, \left\vert{z_2}\right\vert \cos \theta = 0$

Hence by definition:
 * $z_1 \circ z_2 = 0$

Sufficient Condition
Let $z_1 \circ z_2 = 0$.

Then by definition:
 * $\left\vert{z_1}\right\vert \, \left\vert{z_2}\right\vert \cos \theta = 0$

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.