Question

If $z_1$ and $z_2$ satisfy $z+\overline{z}=2|z-1|$ and $\arg (z_1-z_2)=\frac{\pi }{4},$ then $Im(z_1+z_2)=$

Answer 1

Let $z=x+iy$
we have $z+\overline{z}=2|z-1|$
$\Rightarrow \frac{z+\overline{z}}{2}=|z-1|$
$\Rightarrow x=|x+iy-1|\Rightarrow x=|(x-1)+iy|$
$\Rightarrow x^2=(x-1{)}^2+y^2\Rightarrow 2x=1+y^2$
$z_1=x_1+{iy}_1,z_2=x_2+{iy}_2$
then $2x_1=1+y_1^2\dots \left(1\right)$

$2x_2=1+y_2^2\dots \left(2\right)$

Subtract $\left(2\right)$ from $\left(1\right)$ to get,
$2x_1-2x_2=y_1^2-y_2^2=(y_1-y_2)(y_1+y_2)\dots \left(3\right)$
but given that
$\arg (z_1-z_2)=\frac{\pi }{4}$
$\Rightarrow \tan \frac{\pi }{4}=\left|\frac{y_1-y_2}{x_1-x_2}\right|$
As $z_1-z_2$ lies in first quadrant.
So, $x_1-x_2>0,y_1-y_2>0$
$\Rightarrow 1=\frac{y_1-y_2}{x_1-x_2}$
Thus $(y_1-y_2)=(x_1-x_2)\cdots \left(4\right)$
From $\left(3\right)$ and $\left(4\right)$, we get $(y_1+y_2)=2$
Thus, $Im(z_1+z_2)=2$