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 find the imaginary part of $(z_1+z_2)$

Answer 1

Let $z=x+iy$
then $\frac{z+\overline{z}}{2}=x$
$\therefore$ from given relation
$z+\overline{z}=2|z-1|$
$\Rightarrow \frac{z+\overline{z}}{2}=|z-1|$
$\Rightarrow x=|x+iy-1|$
$\Rightarrow x=|(x-1)+iy|$
$\Rightarrow x^2=(x-1{)}^2+y^2$
$\Rightarrow 2x=1+y^2$
If $z_1=x_1+i{y}_1$ and $z_2=x_2+i{y}_2$ then
$2x_1=1+y_1^2$ ......... (1)
and $2x_2=1+y_2^2$ .......... (2)
Subtracting (1) and (2), then
$2(x_1-x_2)=y_1^2-y_2^2$
$2(x_1-x_2)=(y_1+y_2)(y_1-y_2)$ ......... (3)
But given $arg(z_1-z_2)=\pi /4$
$ta{n}^{-1}\left(\frac{y_1-y_2}{x_1-x_2}\right)=\pi /4$
$\Rightarrow \frac{y_1-y_2}{x_1-x_2}=1$
$\therefore y_1-y_2=x_1-x_2$ ....... (4)
$\therefore$ from (3) and (4) we get
$y_1+y_2=2$
$\therefore Im(z_1+z_2)=2$
Ans: 2