Question

| z1 + z2 |^2 = | z1 |^2 + | z2 |^2 then prove that z1/z2 is purely imaginary

Answer 1

Answer:

$\frac{z_1}{z_2}$ is purely imaginary

Step-by-step explanation:

Concept used:

z is purely imaginary if $\bar{z}=-z$

Given:

$|z_1+z_2|^2=|z_1|^2+|z_2|^2$

We know that

$|z_1+z_2|^2=|z_1|^2+|z_2|^2+z_1\bar{z_2}+z_2\bar{z_1}$

$|z_1|^2+|z_2|^2=|z_1|^2+|z_2|^2+z_1\bar{z_2}+z_2\bar{z_1}$

$0=z_1\bar{z_2}+z_2\bar{z_1}$

$z_1\bar{z_2}=-z_2\bar{z_1}$

$\frac{\bar{z_1}}{\bar{z_2}}=-\frac{z_1}{z_2}$

$\bar{(\frac{z_1}{z_2})}=-(\frac{z_1}{z_2})$

Then, $\frac{z_1}{z_2}$ is purely imaginary