If $z_1$ and $z_2$ are two complex numbers and $\arg (\frac{z_1+z_2}{z_1-z_2})=\frac{\pi}{2}$, the figure formed by origin, $z_1$, $z_2$ and$z_1+z_2$

Question

Additional info : $|z_1+z_2|\not = |z_1-z_2|$

I think the diagram is fairly simple to draw.

$z_1+z_2$ and $z_1-z_2$ are perpendicular to each other.

They form a parallogram.

But the answer says it’s a rhombus. But for that to happen $z_1=z_2$ which isn’t mentioned in the question, nor can it be derived. How do I go about it?

Answer 1

$z_1$and $z_2$ are vertices of a parallelogram and $z_1+z_2$ , $ z_1-z_2$ are diagonals. Diagonals are perpendicular, so parallelogram is a rhombus

@Aditya, this is how we do it

$$arg(\frac{z_1+z_2}{z_1-z_2})=\frac{\pi}{2}$$

$\Rightarrow \frac{z_1+z_2}{z_1-z_2}$ is purely imaginary

So, let $\frac{z_1+z_2}{z_1-z_2}=ik$

Use componendo and dividendo, you will get

$\frac{z_1}{z_2}=\frac {ik+1}{ik-1}$

Take modulus both sided you will get $|\frac{z_1}{z_2}|=|\frac {ik+1}{ik-1}|$

$\Rightarrow \frac{|z_1|}{|z_2|}=1$

$\Rightarrow {|z_1|}={|z_2|}$