Complex Numbers 3 Question 1
1. Let $z _1$ and $z _2$ be any two non-zero complex numbers such that $3\left|z _1\right|=4\left|z _2\right|$. If $z=\frac{3 z _1}{2 z _2}+\frac{2 z _2}{3 z _1}$, then
(2019 Main, 10 Jan I)
(a) $|z|=\frac{1}{2} \sqrt{\frac{17}{2}}$
(b) $\operatorname{Im}(z)=0$
(c) $\operatorname{Re}(z)=0$
(d) $|z|=\sqrt{\frac{5}{2}}$
Show Answer
Answer:
Correct Answer: 1. $\left(^{*}\right)$
Solution:
- (*) Given, $3\left|z _1\right|=4\left|z _2\right| \Rightarrow \frac{\left|z _1\right|}{\left|z _2\right|}=\frac{4}{3} \quad\left[\because z _2 \neq 0 \Rightarrow\left|z _2\right| \neq 0\right]$
$\therefore \quad \frac{z _1}{z _2}=\left|\frac{z _1}{z _2}\right| e^{i \theta}$ and $\frac{z _2}{z _1}=\left|\frac{z _2}{z _1}\right| e^{-i \theta}$
$\left[\because z=|z|(\cos \theta+i \sin \theta)=|z| e^{i \theta}\right]$
$\Rightarrow \frac{z _1}{z _2}=\frac{4}{3} e^{i \theta}$ and $\frac{z _2}{z _1}=\frac{3}{4} e^{-i \theta}$
$\Rightarrow \quad \frac{3}{2} \frac{z _1}{z _2}=2 e^{i \theta}$ and $\frac{2}{3} \frac{z _2}{z _1}=\frac{1}{2} e^{-i \theta}$
On adding these two, we get
$$ z=\frac{3}{2} \frac{z _1}{z _2}+\frac{2}{3} \frac{z _2}{z _1}=2 e^{i \theta}+\frac{1}{2} e^{-i \theta} $$
$$ =2 \cos \theta+2 i \sin \theta+\frac{1}{2} \cos \theta-\frac{1}{2} i \sin \theta $$
$\left[\because e^{ \pm i \theta}=(\cos \theta \pm i \sin \theta)\right]$
$$ =\frac{5}{2} \cos \theta+\frac{3}{2} i \sin \theta $$
$\Rightarrow \quad|z|=\sqrt{\frac{5}{2}^{2}+\frac{3}{2}^{2}}=\sqrt{\frac{34}{4}}=\sqrt{\frac{17}{2}}$
Note that $z$ is neither purely imaginary and nor purely real.
‘*’ None of the options is correct.
