so, if z = x+iy,
|z| = \(\sqrt{x^2+y^2}\)
and \( |z|^2 = x^2 + y^2 \)
so if x = 3+bi
|x|^2 = 3^2 + b^2
which is given to be 13
3^2 + b^2 = 13
9+b^2 = 13
b^2 = 13-9
b^2 = 4
b = +\sqrt 4 , -sqrt 4
b = +2,-2
Hence, the possible values of b are +2 or -2.