$$\begin{gathered} \text{ Let }{z}_1\text{ and }{z}_2\text{ be two complex numbers such that }\arg \left(z_1-z_2\right)=\frac{\pi }{4}\text{ and }{z}_1,z_2\text{ satisfy the } \\ \text{ equation }|z-3|=\mathrm{Re}\left(z\right)\text{ Then the imaginary part of }{z}_1+z_2\text{ is equal to } \end{gathered}$$

$\begin{array}{l}{z}_1=x_1+i{y}_1;z_2=x_2+i{y}_2\\ z_1-z_2=\left(x_1-x_2\right)+i\left(y_1-y_2\right)\\ \therefore \arg \left(z_1-z_2\right)=\frac{\pi }{4}\Rightarrow {\tan }^{-1}\left(\frac{y_1-y_2}{x_1-x_2}\right)=\frac{\pi }{4}\\ y_1-y_2=x_1-x_2\dots ..\left(1\right)\\ \left|z_1-3\right|=\mathrm{Re}\left(z_1\right)\Rightarrow {\left(x_1-3\right)}^2+y_1^2=x_1^2\cdots \left(2\right)\\ \left|z_2-3\right|=\mathrm{Re}\left(z_2\right)\Rightarrow {\left({\mathrm{x}}_2-3\right)}^2+y_2^2=x_2^2\\ \end{array}$
$$\begin{gathered} \begin{array}{l}sub \left(2\right) and \left(3\right) \\ {\left({\mathrm{x}}_1-3\right)}^2-{\left(x_2-3\right)}^2+y_1^2-y_2^2=x_1^2-x_2^2\\ \left(x_1-x_2\right)\left(x_1+x_2-6\right)+\left(y_1-y_2\right)\left(y_1+y_2\right)=\left(x_1-x_2\right)\left(x_1+x_2\right)\\ x_1+x_2-6+y_1+y_2=x_1+x_2\Rightarrow y_1+y_2=6\end{array} \end{gathered}$$