Question

If $z_1$ and $z_2$ both satisfy the relation $z+\overline{z}=2|z-1|$ and $arg(z_1-z_2)=\frac{\pi }{4}$, then the imaginary part of $(z_1+z_2)$ is

• A. $0$
• B. $1$
• C. $2$
• D. None of these

Answer 1

Answer: (C)

$2$

Let $z=x+iy$

We have, $z+\overline{z}=2|z-1|$

$\Rightarrow \frac{z+\overline{z}}{2}=|z-1|$

$\Rightarrow x=|x+iy-1|\Rightarrow x=|(x+1)+iy|$

$\Rightarrow x^2={(x-1)}^2+y^2\Rightarrow 2x=1+y^2.$

If $z_1=x_1+iy$ and $z_2=x_2+i{y}_2$

then, $2x_1=1+y_1^2$ ...(1)

and, $2x_1=1+y_2^2$ ...(2)

Subtracting (2) from (1), we get

$2(x_1-x_2)=y_1^2-y_2^2$

$\Rightarrow 2(x_1-x_2)=(y_1-y_2)(y_1-y_2)$ ...(3)

But given arg $(z_1-z_2)=\frac{\pi }{4}$

i.e., ${\tan }^{-1}\left(\frac{y_1-y_2}{x_1-x_2}\right)=\frac{\pi }{4}\Rightarrow \frac{y_1-y_2}{x_1-x_2}=1$

$\therefore y_1-y_2=x_1-x_2$ ...(4)

$\therefore$From (3) and (4) we get

$y_1+y_2=2$

$\therefore Im(z_1+z_2)=2.$