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

The Correct Option is D

\(3\left|z_{1}\right| = 4 \left|z_{2}\right|\)
\(\Rightarrow \frac{\left|z_{1}\right|}{\left|z_{2}\right|} = \frac{4}{3}\)
\(\Rightarrow \frac{\left|3z_{1}\right|}{\left|2z_{2}\right|} = 2\) 
Let \(\frac{3z_{1}}{2z_{2}} = a = 2 \cos \theta + 2 i \sin\theta\)
\(z = \frac{3z_{1}}{2z_{2}} + \frac{2z_{2}}{3z_{1}} = a+ \frac{1}{a}\)
\(= \frac{5}{2} \cos\theta + \frac{3}{2} i \sin \theta\) 
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 \(\frac{3z_{1}}{2z_{2} } = a = \cos\theta + i \sin\theta\)
\(z = \frac{3z_{1} }{2z_{2}} + \frac{2z_{2}}{3z_{1}}\)
\(= a + \frac{1}{a} = 2 \cos\theta\)
\(\therefore Im(z) = 0\)