Question

If ${\theta }_1$ and ${\theta }_2$ are arguments of two non-zero complex numbers $z_1$ and $z_2$ respectively such that $3|z_1|=4|z_2|$. If $z=\frac{3z_1}{2z_2}+\frac{2z_2}{3z_1}$, then

• A. $|z|=\sqrt{\frac{5}{2}}$
• B. $\text{Re}\left(z\right)=\frac{5}{2}\cos ({\theta }_1-{\theta }_2)$
• C. $|z|=\sqrt{\frac{5}{4}}$
• D. $\text{Re}\left(z\right)=\frac{5}{2}\cos ({\theta }_1+{\theta }_2)$

Answer 1

The correct option is B $\text{Re}\left(z\right)=\frac{5}{2}\cos ({\theta }_1-{\theta }_2)$

$z_1=r_1{e}^{i{\theta }_1},z_2=r_2{e}^{i{\theta }_2}$
$\frac{z_1}{z_2}=\frac{r_1{e}^{i{\theta }_1}}{r_2{e}^{i{\theta }_2}}=\frac{r_1}{r_2}{e}^{i({\theta }_1-{\theta }_2)}$
Let ${\theta }_1-{\theta }_2=\alpha$
Since, $3|z_1|=4|z_2|$
$3r_1=4r_2\Rightarrow \frac{r_1}{r_2}=\frac{4}{3}$
$\frac{z_1}{z_2}=\frac{4}{3}{e}^{i\alpha }$
Similarly,
$\frac{z_2}{z_1}=\frac{3}{4}{e}^{-i\alpha }$
Now,
$z=\frac{3z_1}{2z_2}+\frac{2z_2}{3z_1}=\frac{3}{2}\times \frac{4}{3}{e}^{i\alpha }+\frac{2}{3}\times \frac{3}{4}{e}^{-i\alpha }$
$\Rightarrow z=2\cos \alpha +2i\sin \alpha +\frac{1}{2}\cos (-\alpha )+\frac{1}{2}i\sin (-\alpha )$

$\Rightarrow z=\frac{5}{2}\cos \alpha +\frac{3}{2}i\sin \alpha$

$|z|=\sqrt{\frac{25}{4}{\cos }^2\alpha +\frac{9}{4}{\sin }^2\alpha }$
$\text{Im}\left(z\right)=\frac{3}{2}\sin \alpha$
$\text{Re}\left(z\right)=\frac{5}{2}\cos \alpha$