Complex Numbers are Perpendicular iff Dot Product is Zero
Jump to navigation
Jump to search
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.
$\blacksquare$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Dot and Cross Product: $1.$