Question

If $z_1,z_2$ are complex numbers such that $Re\left(z_1\right)=|z_1-1|$, $Re\left(z_2\right)=|z_2-1|$ and $arg(z_1-z_2)=\frac{\pi }{6}$, then $Im(z_1+z_2)$ is equal to:

• A. $2\sqrt{3}$
• B. $\frac{2}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is A $2\sqrt{3}$

Let $z_1=x_1+i{y}_1$ and $z_2=x_2+i{y}_2$
$Re\left(z_1\right)=|z_1-1|$
$\Rightarrow x_1^2={(x_1-1)}^2+y_1^2$
$\Rightarrow y_1^2-2x_1+1=0\cdots \left(1\right)$

$Re\left(z_2\right)=|z_2-1|$
$\Rightarrow x_2^2={(x_2-1)}^2+y_2^2$
$\Rightarrow y_2^2-2x_2+1=0\cdots \left(2\right)$
Subtracting equation $\left(2\right)$ from $\left(1\right)$, we get
$(y_1^2-y_2^2)-2(x_1-x_2)=0$
$\Rightarrow (y_1+y_2)(y_1-y_2)=2(x_1-x_2)$
$\Rightarrow y_1+y_2=2\left(\frac{x_1-x_2}{y_1-y_2}\right)$

$arg(z_1-z_2)=\frac{\pi }{6}$
$\Rightarrow {\tan }^{-1}\left(\frac{y_1-y_2}{x_1-x_2}\right)=\frac{\pi }{6}$
$\Rightarrow \frac{y_1-y_2}{x_1-x_2}=\frac{1}{\sqrt{3}}$
$\therefore y_1+y_2=2\sqrt{3}$