Modulus of Sum equals Modulus of Distance implies Quotient is Imaginary

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Theorem

Let $z_1$ and $z_2$ be complex numbers such that:

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


Then $\dfrac {z_2} {z_1}$ is wholly imaginary.


Corollary

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 $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$.


Then:

\(\ds \cmod {z_1 + z_2}\) \(=\) \(\ds \cmod {z_1 - z_2}\)
\(\ds \leadsto \ \ \) \(\ds \paren {x_1 + x_2}^2 + \paren {y_1 + y_2}^2\) \(=\) \(\ds \paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2\) Definition of Complex Modulus
\(\ds \leadsto \ \ \) \(\ds {x_1}^2 + 2 x_1 x_2 + {x_2}^2 + {y_1}^2 + 2 y_1 y_2 + {y_2}^2\) \(=\) \(\ds {x_1}^2 - 2 x_1 x_2 + {x_1}^2 + {y_1}^2 - 2 y_1 y_2 + {y_1}^2\) Square of Sum, Square of Difference
\(\ds \leadsto \ \ \) \(\ds 4 x_1 x_2 + 4 y_1 y_2\) \(=\) \(\ds 0\) simplifying
\(\ds \leadsto \ \ \) \(\ds x_1 x_2 + y_1 y_2\) \(=\) \(\ds 0\) simplifying


Now we have:

\(\ds \dfrac {z_1} {z_2}\) \(=\) \(\ds \frac {x_1 + i y_1} {x_2 + i y_2}\)
\(\ds \) \(=\) \(\ds \frac {\paren {x_1 + i y_1} \paren {x_2 - i y_2} } { {x_2}^2 + {y_2}^2}\) Definition of Complex Division
\(\ds \) \(=\) \(\ds \frac {x_1 x_2 + y_1 y_2} { {x_2}^2 + {y_2}^2} + \frac {i \paren {x_2 y_1 - x_1 y_2} } { {x_2}^2 + {y_2}^2}\) Definition of Complex Multiplication


But we have:

$x_1 x_2 + y_1 y_2 = 0$

Thus:

$\dfrac {z_1} {z_2} = \dfrac {i \paren {x_2 y_1 - x_1 y_2} } { {x_2}^2 + {y_2}^2}$

which is wholly imaginary.

$\blacksquare$


Sources

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