# Let z1 and z2 be any two non-zero complex numbers such that

Question:

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:

1. (1) $3 / 2 \leq|z| \leq 5 / 2$

2. (2) $1 / 2 \leq|z| \leq 5 / 2$

3. (3) $3 / 2 \leq|z| \leq 7 / 2$

4. (4) $1 \leq|z| \leq 5 / 2$

Correct Option: 1

Solution:

Let $z_{1}=r_{1} e^{i \theta}$ and $z_{2}=r_{2} e^{i \phi}$

$3\left|z_{1}\right|=4\left|z_{2}\right| \Rightarrow 3 r_{1}=4 r_{2}$

$z=\frac{3 z_{1}}{2 z_{2}}+\frac{2 z_{2}}{3 z_{1}}=\frac{3 r_{1}}{2 r_{2}} e^{i(\theta-\phi)}+\frac{2}{3} \frac{r_{2}}{r_{1}} e^{i(\phi-\theta)}$

$=\frac{3}{2} \times \frac{4}{3}(\cos (\theta-\phi)+i \sin (\theta-\phi))$

$+\frac{2}{3} \times \frac{3}{4}[\cos (\theta-\phi)-i \sin (\theta-\phi)]$

$z=\left(2+\frac{1}{2}\right) \cos (\theta-\phi)+\mathrm{i}\left(2-\frac{1}{2}\right) \sin (\theta-\phi)$

$\therefore|z|=\sqrt{\frac{25}{4} \cos ^{2}(\theta-\phi)+\frac{9}{4} \sin ^{2}(\theta-\phi)}$

$=\sqrt{\frac{16}{4} \cos ^{2}(\theta-\phi)+\frac{9}{4}}$

$\Rightarrow \frac{3}{2} \leq|z| \leq \frac{5}{2}$