Mathematics - Complex Number Question with Solution | TestHub

Let $z_{1 }=x_1+i{y}_1$ and $z_2=x_2+i{y}_2$

Given $\left|z_1-1\right|=Re\left(z_1\right)$

$\Rightarrow {\left(x_1-1\right)}^2+y_1^2=x_1^2$

$\Rightarrow y_1^2-2x_1+1=0 ..............\left(1\right)$

$\left|z_2-1\right|=Re\left(z_2\right)$

$\Rightarrow {\left(x_2-1\right)}^2+y_2^2=x_2^2$

$\Rightarrow y_2^2 -2x_2+1=0 ...........\left(2\right)$

Subtracting equation $\left(2\right)$ from equation $\left(1\right)$, we get

$y_1^2-y_2^2-2\left(x_1-x_2\right)=0$

$\Rightarrow \left(y_1-y_2\right) \left(y_1+y_2\right)=2\left(x_1-x_2\right)$

$\Rightarrow y_1+y_2=2\left(\frac{x_1-x_2}{y_1-y_2}\right) ............\left(3\right)$

Also given, $arg \left(z_1-z_2\right)=\frac{\pi }{6}$

$\Rightarrow {\tan }^{-1}\left(\frac{y_1-y_2}{x_1-x_2}\right)=\frac{\pi }{6}$

$\Rightarrow \frac{y_1-y_2}{x_1-x_2}=\frac{1}{\sqrt{3}} .........\left(4\right)$

Putting equation $\left(4\right)$ in equation $\left(3\right)$, we get

$y_1+y_2=2\sqrt{3}\Rightarrow Im\left(z_1+z_2\right)=2\sqrt{3}$