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

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$