Question

Let $z_1$ and $z_2$ be any two non-zero complex numbers such that $3|z_1|=4|z_2|$. If $z=\frac{3z_1}{2z_2}+\frac{2z_2}{3z_1}$ then :

• A. $\text{Im}\left(z\right)=0$
• B. $\text{Re}\left(z\right)=0$
• C. $|z|=\frac{1}{2}\sqrt{\frac{17}{2}}$
• D. $|z|=\sqrt{\frac{5}{2}}$
• E. $\text{Re}\left(z\right)=\frac{5}{2}\cos ({\theta }_1-{\theta }_2)$

Answer 1

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

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

$\Rightarrow z=\frac{5}{2}\cos ({\theta }_1-{\theta }_2)+\frac{3}{2}i\sin ({\theta }_1-{\theta }_2)$

$|z|=\sqrt{\frac{25}{4}{\cos }^2({\theta }_1-{\theta }_2)+\frac{9}{4}{\sin }^2({\theta }_1-{\theta }_2)}$
$\text{Im}\left(z\right)=\frac{3}{2}\sin ({\theta }_1-{\theta }_2)$
$\text{Re}\left(z\right)=\frac{5}{2}\cos ({\theta }_1-{\theta }_2)$

No options matches with the answer,
It is bonus question.