Consider a complex equation, z=x+iy It conjugate is z=x−iy Now (7+i)(z+z)−(4+i)(z−z)+116i‌‌=0 (7+i)(2x)−(8iy−2y)+116i‌‌=0 (14x+2y)+i(2x−8y+116)‌‌=0 The equations obtained are 2x−8y+116‌‌=0 14x+2y‌‌=0 From above two equations, x=−2,y=14 The modulus is, |z|‌‌=√x2+y2 ‌‌=√(−2)2+142 ‌‌=√200 ‌ Thus, ‌zz=|z|2=(√200)2‌‌=200