Modulus of Sum equals Modulus of Distance implies Quotient is Imaginary/Corollary
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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: $147 \ \text{(b)}$