Let $z_{1}$ and $z_{2}$ be two complex numbers such that $\arg \left(\mathrm{z}_{1}-\mathrm{z}_{2}\right)=\frac{\pi}{4}$ and $\mathrm{z}_{1}, \mathrm{z}_{2}$ satisfy the equation $|z-3|=\operatorname{Re}(z)$. Then the imaginary part of $z_{1}+z_{2}$ is equal to
$|z-3|=\operatorname{Re}(z)$
let $Z=x=$ iy
$\Rightarrow(x-3)^{2}+y^{2}=x^{2}$
$\Rightarrow x^{2}+9-6 x+y^{2}=x^{2}$
$\Rightarrow y^{2}=6 x-9$
$\Rightarrow y^{2}=6\left(x-\frac{3}{2}\right)$
$\Rightarrow z_{1}$ and $z_{2}$ lie on the parabola mentioned in eq.(1)
$\arg \left(\mathrm{z}_{1}-\mathrm{z}_{2}\right)=\frac{\pi}{4}$
$\Rightarrow$ Slope of $\mathrm{PQ}=1$
Let $\mathrm{P}\left(\frac{3}{2}+\frac{3}{2} \mathrm{t}_{1}^{2}, 3 \mathrm{t}_{1}\right)$ and $\mathrm{Q}\left(\frac{3}{2}+\frac{3}{2} \mathrm{t}_{2}^{2}, 3 \mathrm{t}_{2}\right)$
Slope of PQ $=\frac{3\left(t_{2}-t_{1}\right)}{\frac{3}{2}\left(t_{1}^{2}-t_{1}^{2}\right)}=1$
$\Rightarrow \frac{2}{t_{2}+t_{2}}=1$
$\Rightarrow t_{2}+t_{1}=2$
$\operatorname{Im}\left(\mathrm{z}_{1}+\mathrm{z}_{2}\right)=3 \mathrm{t}_{1}+3 \mathrm{t}_{2}=3\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)=3(2)$
Ans. $6.00$