It is given that, f(z)=z4+z3+2z2+4z−8 have a root 2i, so one more root will be −2i, so (z2+4) is the factor of z4+z3+2z2+4z−8. So, z4+z3+2z2+4z−8 =(z2+4)(z2+z−2) and z2+z−2=(z+2)(z−1) Therefore, the roots of f(z) are 2i,−2i,−2 and 1 .