Question

It is given that complex numbers $z_1$ and $z_2$ satisfy $\left|z_1\right|=2$ and $\left|z_2\right|=3.$ If the included angle of their corresponding vectors is $60^{\circ }$ then $\left|\frac{z_1+z_2}{z_1-z_2}\right|$ can be expressed on $\frac{\sqrt{N}}{7}$ where $N$ is natural number then $N$ equals

Answer 1

Answer: 133

Using cosine rule
$\left|z_1+z_2\right|=\sqrt{{\left|z_1\right|}^2+{\left|z_2\right|}^2-2\left|z_1\right|\left|z_2\right|\cos 120^{\circ }}$
$=\sqrt{4+9-2\cdot 3}=\sqrt{19}$

and $\left|z_1-z_2\right|=\sqrt{{\left|z_1\right|}^2+{\left|z_2\right|}^2-2\left|z_1\right|\left|z_1\right|\cos 60^{\circ }}$
$=\sqrt{4+9-2\cdot 3}=\sqrt{7}$
$\therefore \left|\frac{z_1+z_2}{z_1+z_2}\right|=\sqrt{\frac{19}{7}}=\frac{\sqrt{133}}{7}$
$\Rightarrow N=133$