Question

If $\frac{4z_1}{3z_2}$ is a purely imaginary number other than zero then find the value of $\frac{4z_1}{3z_2}$

Answer 1

Let $z_1=x_1+i{y}_1$ and $z_2=x_2+i{y}_2$

Now, $\frac{4z_1}{3z_2}$ $=\frac{4z_1\overline{z_2}}{3z_2\overline{z_2}}=\frac{4(x_1+iy)(x_2-i{y}_2)}{3|z_2{|}^2}$ $[\because z_1\overline{z}=|z{|}^2]$

$=\frac{4}{3|z_2{|}^2}$ $\{x_1{x}_2-i{x}_1{y}_2+i{x}_2{y}_1+y_1{y}_2\}$

$=\frac{4}{3|z_2{|}^2}\left\{\right(x_1{x}_2+y_1{y}_2)+I(x_2{y}_1-x_1{y}_2\left)\right\}$

$=\frac{4}{3|z_2{|}^2}(x_1{x}_2+y_1{y}_2)+\frac{4i}{3|z_2{|}^2}(x_2{y}_1-x_1{y}_2)$ …….$\left(1\right)$

Given the $\frac{4z_1}{3z_2}$ is a purely imaginary number

$\therefore \frac{4}{3|z_2{|}^2}(x_1{x}_2+y_1{y}_2)=0$

$\Rightarrow x_1{x}_2+y_1{y}_2=0$

$\Rightarrow x_1{x}_2=-y_2{y}_2$

$\Rightarrow x_2=dfrac-y_1{y}_2{x}_1$ …………$\left(2\right)$

In $\left(1\right)$, substituting $x_2$ from $\left(1\right)$ we get

$\frac{4z_1}{3z_2}=\frac{4i}{3|z_2{|}^2}(\frac{-y_1{y}_2}{x_1}\cdot y_1-x_1{y}_2)=\frac{4i}{3|z_2{|}^2}\left\{\frac{-y_1^2{y}_2-x_1^2{y}_2}{x_1}\right\}$

$=\frac{-4i}{3|z_2{|}^2{x}_1}$ $\left(y_1^2{y}_2{x}_1^2{y}_2\right)$

$=\frac{-4y_{2}i}{3|z_2{|}^2{x}_1}(x_1^2+y_1^2)$

$=\frac{-4y_{2}i}{3|z_2{|}^2{x}_1|\cdot |z_1{|}^2}$ $[\because |z_1|=\sqrt{x_1^2+y_1^2},|z_1{|}^2=x_1^2{y}_1^2]$

$=\frac{-4|z_1{|}^2{y}_2}{3|z_2{|}^2{x}_1}i$

$=\frac{-4|z_1{|}^{2}Im\left(z_2\right)i}{3|z_2{|}^{2}Re\left(z_1\right)}$.