Modulus of Sum equals Modulus of Distance implies Quotient is Imaginary/Corollary

Corollary to Modulus of Sum equals Modulus of Distance implies Quotient is Imaginary

Let $P_1$ and $P_2$ be points embedded in the complex plane.

Let $P_1$ and $P_2$ be represented by the complex numbers $z_1$ and $z_2$ be complex numbers such that:

$\cmod {z_1 + z_2} = \cmod {z_1 - z_2}$

Then:

$\angle P_1 O P_2 = 90 \degrees$

Proof

Let $z_1 = x_1 + i y_1 = r_1 e^{i \theta_1}, z_2 = x_2 + i y_2 = r_2 e^{i \theta_2}$.

We have from Division of Complex Numbers in Polar Form that:

$\dfrac {z_1} {z_2} = \dfrac {r_1} {r_2} \paren {\map \cos {\theta_1 - \theta_2} + i \, \map \sin {\theta_1 - \theta_2} }$

From Modulus of Sum equals Modulus of Distance implies Quotient is Imaginary, $\dfrac {z_1} {z_2}$ is wholly imaginary.

Hence $\map \cos {\theta_1 - \theta_2} = 0$.

Hence from Cosine of Half-Integer Multiple of Pi, $\theta_1 - \theta_2 = \paren {n + \dfrac 1 2} \pi$ for some $n$.

In the geometrical context, that means:

$\theta_1 - \theta_2 = 90 \degrees$

$\blacksquare$

Sources

• 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $147 \ \text{(b)}$