If a complex number z is such that ( 7 + i ) ( z + z ¯ ) - ( 4 + i ) ( z - z ¯ ) + 116 i = 0 ,…

Consider a complex equation, $z=x+i y$

Its conjugate is $\overline{z}=x-iy$

Now,$\begin{array}{r}(7+i)(z+\overline{z})-(4+i)(z-\overline{z})+116i=0\end{array}$

$\Rightarrow (7+i)\left(2x\right)-(8iy-2y)+116i=0$

$\Rightarrow (14x+2y)+i(2x-8y+116)=0$

The equations obtained are

$\Rightarrow 2x-8y+116=0$

$\Rightarrow 14x+2y=0$

From the above two equations,

$x=-2, y=14$

The modulus is

$\left|z\right|=\sqrt{x^2+y^2}$

$=\sqrt{(-2{)}^2+{14}^2}$

$=\sqrt{200}$

Thus, $z\overline{z}=|z{|}^2=(\sqrt{200}{)}^2=200$