Question

It is given that complex nunbers $z_1$ and $z_2$ satisfy $\left|z_1\right|=2$ and $\left|z_2\right|=3$. If the included angle of their corresponding vectors is ${60}^o$, then find the value of $\left|\frac{z_1+z_2}{z_1-z_2}\right|$.

• A. $\frac{\sqrt{19}}{\sqrt{7}}$
• B. $\frac{\sqrt{7}}{\sqrt{19}}$
• C. $\frac{\sqrt{12}}{\sqrt{19}}$
• D. $\frac{\sqrt{19}}{\sqrt{12}}$

Answer 1

The correct option is A $\frac{\sqrt{19}}{\sqrt{7}}$

Let $z_1=2(\cos \theta +i\sin \theta )⟹z_2=3\left(\right(\cos (\theta +{60}^o)+i\sin (\theta +{60}^o))$
$⟹\frac{z_2}{z_1}=\frac{3\left(\cos \right(\theta +{60}^o)+i\sin (\theta +{60}^o\left)\right)}{2(\cos \theta +i\sin \theta )}$
$\frac{z_2}{z_1}=\frac{3}{2}(\cos {60}^o+i\sin {60}^o)=\frac{3}{2}(\frac{1}{2}+i\frac{\sqrt{2}}{2})$
$\therefore \left|\frac{z_1+z_2}{z_1-z_2}\right|=\left|\frac{1+\frac{z_2}{z_1}}{1-\frac{z_2}{z_1}}\right|$
$=\left|\frac{1+\frac{3}{2}(\frac{1}{2}+i\frac{\sqrt{3}}{2})}{1-\frac{3}{2}(\frac{1}{2}+i\frac{\sqrt{3}}{2})}\right|$
$=\left|\frac{7+i3\sqrt{3}}{1-i3\sqrt{3}}\right|$
$=\frac{\sqrt{49+27}}{\sqrt{1+27}}=\frac{\sqrt{76}}{\sqrt{28}}=\frac{\sqrt{19}}{\sqrt{7}}$
Ans: A