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. $\left|z\right|=\frac{1}{2}\sqrt{\frac{17}{2}}$
• B. $Re(z)=0$
• C. $\left|z\right|=\sqrt{\frac{5}{2}}$
• D. $Im(z)=0$

Answer 1

Answer: (D)

$3\left|z_1\right|=4\left|z_2\right|$
$\Rightarrow \frac{|z_1|}{|z_2|}=\frac{4}{3}$
$\Rightarrow \frac{|3z_1|}{|2z_2|}=2$
Let $\frac{3z_1}{2z_2}=a=2\cos \theta +2i\sin \theta$
$z=\frac{3z_1}{2z_2}+\frac{2z_2}{3z_1}=a+\frac{1}{a}$
$=\frac{5}{2}\cos \theta +\frac{3}{2}i\sin \theta$
Now all options are incorrect .
There is a misprint in the problem actual problem should be :
"Let $z_1$ and $z_2$ be any non-zero complex number such that $3|z_1|=2|z_2|$.
If $z=\frac{3z_1}{2z_2}+\frac{2z_2}{3z_1}$ , then "
Given
$3|z_1|=2|z_2|$
Now $\left|\frac{3z_1}{2z_2}\right|=1$
Let $\frac{3z_1}{2z_2}=a=\cos \theta +i\sin \theta$
$z=\frac{3z_1}{2z_2}+\frac{2z_2}{3z_1}$
$=a+\frac{1}{a}=2\cos \theta$
$\therefore Im(z)=0$