Let z₁ and z₂ be two complex numbers such that $\arg ({z_1} - {z_2}) = {\pi \over 4}$ and z₁, z₂ satisfy the equation | z $-$ 3 | = Re(z). Then the imaginary part of z₁ + z₂ is equal to ___________.

Let z₁ = x₁ + iy ; z₂ = x₂ + iy₂<br><br>z₁ $-$ z₂ = (x₁ $-$ x₂) + i(y₁ $-$ y₂)<br><br>$\therefore$ $\arg ({z_1} - {z_2}) = {\pi \over 4}$ $\Rightarrow$ $${\tan ^{ - 1}}\left( {{{{y_1} - {y_2}} \over {{x_1} - {x_2}}}} \right) = {\pi \over 4}$$<br><br>${y_1} - {y_2} = {x_1} - {x_2}$ ....... (1)<br><br>$$|{z_1} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_1}) \Rightarrow {({x_1} - 3)^2} + {y_1}^2 = {x_1}^2$$ .... (2)<br><br>$$|{z_2} - 3|\, = {\mathop{\rm Re}\nolimits} ({z_2}) \Rightarrow {({x_2} - 3)^2} + {y_2}^2 = {x_2}^2$$ .... (3)<br><br>sub (2) &amp; (3)<br><br>${({x_1} - 3)^2} - {({x_2} - 3)^2} + {y_1}^2 - {y_2}^2 = {x_1}^2 - {x_2}^2$<br><br>$({x_1} - {x_2})({x_1} + {x_2} - 6) + ({y_1} - {y_2})({y_1} + {y_2})$<br><br>$= ({x_1} - {x_2})({x_1} + {x_2})$<br><br>${x_1} + {x_2} - 6 + {y_1} + {y_2} = {x_1} + {x_2} \Rightarrow {y_1} + {y_2} = 6$