If ${\left|z_1+z_2\right|}^2={\left|z_1\right|}^2+{\left|z_2\right|}^2$, where $z_1$ and $z_2$ are two complex numbers, then

Given that: ${\left|z_1+z_2\right|}^2={\left|z_1\right|}^2+{\left|z_2\right|}^2$, where $z_1$ and $z_2$ are two complex numbers

${\left|z_1+z_2\right|}^2={\left|z_1\right|}^2+{\left|z_2\right|}^2+2\left|z_1\right|\left|z_2\right|\cos (x-y)$, where $x \& y$ are arguments of $z_1 \& z_2$.

Here, $2\left|z_1\right|\left|z_2\right|\cos (x-y)=0$

$\Rightarrow \cos (x-y)=0$

$\Rightarrow x-y=(2n-1)\frac{\mathrm{\pi }}{2} \text{where} n\in I$

Now, $arg\left(\frac{z_1}{z_2}\right)=arg\left(z_1\right)-arg\left(z_2\right)=x-y=(2n-1)\frac{\mathrm{\pi }}{2} \text{where} n\in I$

$\Rightarrow \frac{z_1}{z_2}$ is a purely imaginary.