If $z_{1},\ z_{2}$ are two complex numbers such that $arg\left( { z }_{ 1 }+{ z }_{ 2 } \right) =0$ and $Im\left( { z }_{ 1 }{ z }_{ 2 } \right) =0$, then

Let the complex numbers be,

$z_1=a+ib,z_2=x+iy$

$arg(z_1+z_2)=arg(a+x+i(b+y))=0$

$\tan^{-1}\left ( \dfrac{b+y}{a+x} \right )=0$

$\dfrac{b+y}{a+x}=0$

$b+y=0$.....(1)

$Im(z_1z_2)=Im(ax-by+i(ay+bx))=0$

$(ay+bx)=0$

from (1),

$ay=xy\Rightarrow a=x$

$\Rightarrow b=-y,a=x$

$\therefore z_1=x-iy,z_2=x+iy$

Hence proved, $z_1=\bar{z}_2$