Question

The complex numbers $z_1$ and $z_2$ satisfy $|z_1|=2,|z_2|=3$. If the included angle of the corresponding vectors is ${60}^o$ then $\left|\frac{z_1+z_2}{z_1-z_2}\right|$ is :

• A. $\frac{\sqrt{133}}{7}$
• B. $\frac{7}{\sqrt{133}}$
• C. $\frac{6}{\sqrt{130}}$
• D. $\frac{\sqrt{130}}{6}$

Answer 1

The correct option is A $\frac{\sqrt{133}}{7}$

$\left|z_1\right|=2$

$\left|z_2\right|=3$

$\theta =60°$

Now,

${|z_1+z_2|}^2={\left|z_1\right)}^2+{\left|z_2\right|}^2+2\left|z_1\right|\left|z_2\right|\cos \theta$

$\Rightarrow {|z_1+z_2|}^2=2^2+3^2+2\times 2\times 3\times \frac{1}{2}$

$\Rightarrow |z_1+z_2|=\sqrt{4+9+6}=\sqrt{19}$

Similarly,

${|z_1-z_2|}^2={\left|z_1\right)}^2+{\left|z_2\right|}^2-2\left|z_1\right|\left|z_2\right|\cos \theta$

$\Rightarrow {|z_1-z_2|}^2=2^2+3^2-2\times 2\times 3\times \frac{1}{2}$

$\Rightarrow |z_1-z_2|=\sqrt{4+9-6}=\sqrt{7}$

Therefore,

$\frac{|z_1+z_2|}{|z_1-z_2|}=\frac{\sqrt{19}}{\sqrt{7}}=\frac{\sqrt{19}}{\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{133}}{7}$