If $|{z_1+z_2}|>|z_1-z_2|$ then prove that $-\frac{\pi}{2}<arg\frac{z_1}{z_2}<\frac{\pi}{2}$
Here is my progress:
squaring on both sides of the given inequality, and using the property of complex numbers:
let z be any complex number then $|z|^2=z .\bar{z}$ , where $\bar{z}$ denotes the conjugate of z
so, we get:
$(z_1+z_2)(\bar{z_1}+\bar{z_2})> (z_1-z_2)(\bar{z_1}-\bar{z_2})$
simplifying,
$2z_1\bar{z_2}+2z_2\bar{z_1}>0$
$\frac{z_1}{\bar{z_2}}>-\frac{z_2}{\bar{z_2}}$
Using the euler form and assuming argument of $z_1$ and $z_2$ to be $\alpha$ and $\beta$ respectively, we get after simplifying:
$e^{2i\alpha}>-e^{2i\beta}$
solving further we get:
$e^{2i(\alpha-\beta)}>e^{-i\pi}$
Hence,
$\alpha-\beta>-\frac{\pi}{2}$
which gives us half of the part of the inequality that we need to prove (as $\alpha-\beta$ is the argument of $\frac{z_1}{z_2}$) But how would we prove the other part of the inequality?