KEAM 2023 Math Solved Paper

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