f(z)=zn+an−1zn−1+...+a1z+a0 has no real root We know that complex roots occur in conjugate pairs Consider f(z)=1+z=0 f(z)=1+z+z2=0 and f(s)=1−2+22−2=0 1+z=0 has one real root 1−z+z2−z3=0 has real solution z=1 1+z+z2=0 has two imaginary roots ω,ω2. From this we shall conclude that to get imaginary roots the degree of f(z) must be even