Question

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

Answer 1

Let $z=x+iy,z_1=x_1+i{y}_1$ and $z_2=x_2+i{y}_2$

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

$\Rightarrow (x+iy)+(x-iy)=2|x-1+iy|\Rightarrow 2x=2\sqrt{(x-1{)}^2+y^2}$

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

$\Rightarrow 2x=1+y^2$ ...(i)

Since, $z_1$ and $z_2$ both satisfy Eq. (i),

$\therefore 2x_1=1+{y_1}^2$ and $2x_2=1+{y_2}^2$

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

Again, $z_1-z_2=(x_1-x_2)+i(y_1-y_2)$

$\therefore tan\theta =\frac{y_1-y_2}{x_1-x_2}$, where $\theta =arg(z_1-z_2)$

$\Rightarrow tan\frac{\pi }{4}=\frac{y_1-y_2}{x_1-x_2}$

$\Rightarrow 1=\frac{y_1-y_2}{x_1-x_2}$

From Eq. (ii), we get

$2=y_1+y_2$.

i.e., $m(z_1+z_2)=2$