If f(z)=zⁿ+aₙ-1}zⁿ-1}+·s+a₁zⁿ+a₀ {R}[z] is a polynomial in z with no root over {R}. then (f) is

Correct Answer is : an even number

Correct Answer is : an even number

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KEAM 2023 Math Solved Paper

Question : 19 of 120

Marks: +1, -0

f(z)=z^n+a_{n-1}z^{n-1}+\cdots+a_1z^n+a_0 \in \mathbb{R}[z]

z

\mathbb{R}

\deg(f)

Solution:

f(z)=z^n+a_{n-1}z^{n-1}+\dots+a_1z+a_0

has

no real root

We know that complex roots occur in conjugate pairs

Consider

f(z)=1+z=0

f(z)=1+z+z^2=0

and

f(s)=1-2+2^2-2=0

1+z=0

has one real root

1-z+z^2-z^3=0

has real solution

z=1

1+z+z^2=0

has two imaginary roots

\omega, \omega^2

. From this we shall conclude that

to get imaginary roots the degree of

f(z)

must be even

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