Question

If $z_{1}$ and $z_{2}$ both satisfy $z+\bar{z}=2\left|z-1\right|$ , arg $\left(z_{1}-z_{2}\right)=\frac{\pi}{4}$ , then find Im $\left(z_{1}+z_{2}\right)$ .

• A. $4$
• B. $2$
• C. $1$
• D. $3$

Answer 1

Answer: (B)

Let $z = x + iy$ ,
$z_{1} =x_{1} +iy_{1}$
and $z_{2}=x_{2}+iy_{2}$,
$z+\bar{z}=2\left|z-1\right|$
$\Rightarrow \left(x+iy\right)+\left(x-iy\right)$
$=2\left|x-1+iy\right|$
$\Rightarrow \, 2x=1+y^{2}\quad...\left(i\right)$
Since $z_{1}$ and $z_{2}$ both satisfy $\left(i\right)$, we have
$2x_{1}=1+y_{1}^{2}$ and $2x_{2}=1+y_{2}^{2}$
$\Rightarrow \, 2\left(x_{1}-x_{2}\right)=\left(y_{1}+y_{2}\right)\left(y_{1}-y_{2}\right)$
$\Rightarrow \, 2=\left(y_{1}+y_{2}\right)\left(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\right)\quad\ldots\left(ii\right)$
Again $z_{1}-z_{2}=\left(x_{1}-x_{2}\right)+i\left(y_{1}-y_{2}\right)$
Therefore, $tan\, \theta=\frac{y _{1}-y_{2}}{x_{1}-x_{2}}$ ,
where $\theta=arg\left(z_{1}-z_{2}\right)$
$\Rightarrow \, tan \frac{\pi}{4}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\quad$ (since $\theta\frac{\pi}{4}$)
i.e., $1=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$
From $\left(ii\right)$ , we get $2=y_{1}+y_{2}$ ,
i.e., Im $\left(z_{1}+z_{2}\right)=2$