For 2

a complex number z = 0 if and only if Re(z) = 0 and Im(z) =0

(x^2 - y^2) + i (2xy -18) = 0

x^2 = y^2 x =+y

and 2xy = 18

If x= y 2x^2 = 18 x=+3

If x = -y -2x^2 =18 and there are no solutions

only solutions are (3,3) and (-3,-3)

For 1)

(x+2i)(x-2i) = x^2 + 2ix - 2ix - 4i^2 = x^2 + 4

also

Even if

how do you conclude 4x+12 = 9 -x^2 ?

you should have x^2 + 4 = 5 -12 i

x^2 = 1-12i

if x is real x^2 must be real so there are no solutions

yes i^2 =-1 and -i^2 = 1