Two nonzero complex numbers such that $| z_1 + z_2 |$ = $| z_1 - z_2 |$. If $z_2 ≠ 0$, show that $\frac{z_1}{z_2}$ is a pure imaginary.

Question

Two nonzero complex numbers are such that $| z_1 + z_2 |$ = $| z_1 - z_2 |$. If $z_2 ≠ 0$, show that $\frac{z_1}{z_2}$ is a pure imaginary.

I'm stuck with this problem, I have be using $arg(\frac{z_1}{z_2})$ = $arg(z_1) - arg(z_2)$ but so far I didn get the answer

Can you help me?

Answer 1

$\begin{align}|z_1+z_2|=|z_1-z_2| &\implies |z_1+z_2|^2=|z_1-z_2|^2\\ &\implies (z_1+z_2)\overline{(z_1+z_2)}=(z_1-z_2)\overline{(z_1-z_2)}\\ &\implies|z_1|^2+|z_2|^2+z_1\overline{z_2}+ z_2\overline{z_1}= |z_1|^2+|z_2|^2-z_1\overline{z_2}- z_2\overline{z_1}\\ &\implies z_1\overline{z_2}+ z_2\overline{z_1}=0\\ &\implies \frac{z_1}{z_2}+\overline{\left(\frac{z_1}{z_2}\right)}=0\\ &\implies \Re(\frac{z_1}{z_2})=0 \end{align} $

Answer 2

Let $z_1 = a_1 + i b_1$, $z_2 = a_2 + i b_2$, with $a_i, b_i \in \mathbb{R} $. Then, we have by assumption: $$ (a_1+a_2)^2 + (b_1 + b_2)^2 = (a_1 - a_2)^2 + (b_1 - b_2)^2 $$ After simplification, we get: $$ 4 a_1 a_2 + 4 b_1 b_2 $$ that is: $$ a_1 a_2 = -b_1 b_2 $$ Moreover, we know that: $$ \frac{z_1}{z_2}= \frac{(a_1 + i b_1)(a_2 - i b_2)}{ a_2 ^2 + b_2 ^2} = \frac{a_1 a_2 - i a_1 b_2 + i b_1 a_2 + b_1 b_2}{a_2 ^2 + b_2 ^2 } $$ Using the previous relation, we obtain: $$ \frac{z_1}{z_2}=\frac{ b_1 a_2 - a_1 b_2 }{a_2 ^2 + b_2 ^2 } i $$ which is clearly pure imaginary.

Answer 3

Geometrically are $0,z_1,z_1+z_2$ and $z_2$ vertices of a parallelogram.

$|z_1-z_2|=|z_1+z_2|$ signifies that the diagonals have equal length. This signifies that the parallelogram is a rectangle (incl. a square as a special case.

Orthogonal adjacent sides are represented by complex numbers whose arguments differ in $\pm \pi\over 2,$ which completes the proof.

Answer 4

Let $w=\frac{z_1}{z_2}$. Then, from $| z_1 + z_2 | =| z_1 - z_2 |$, we get

$$|w+1|= |w-1|$$

i.e. $w$ is an equal distance line from the points $-1$ and $1$, thus representing the imaginary axis.