Put \(z=x+iy\). Multiply by the conjugate of the denominator. Its imaginary numerator is \[xy-(x+4)(y+4)=-4(x+y+4).\] Thus \(x+y=-4\). The modulus gives \[x^2+(-4-x)^2=10\Rightarrow x^2+4x+3=0\Rightarrow x=-3,-1.\] Therefore \[z=-3-i\quad\text{or}\quad z=-1-3i.\] Neither has \(z+4i=0\), and both give a real original quotient.