JEE Advanced 2016 Paper 2 · Mathematics · Complex Numbers · medium

Let ω=e^2π i/7, α=ω+ω^2+ω^4 and β=ω^3+ω^5+ω^6. The quadratic equation with real coefficients whose roots are α^2 and β^2 is

Single correct option

Let

\omega=e^{2\pi i/7}

\alpha=\omega+\omega^{2}+\omega^{4}

and

\beta=\omega^{3}+\omega^{5}+\omega^{6}

. The quadratic equation with real coefficients whose roots are

\alpha^{2}

and

\beta^{2}

Show answer and solution

Answer: (C) $x^{2}+3x+4=0$

Since

1+\omega+\dots+\omega^{6}=0

, we get

\alpha+\beta=-1

. Also

\alpha\beta=\sum

of nine terms

\omega^{a+b}

with

a\in\{1,2,4\}

b\in\{3,5,6\}

; the exponents reduce mod 7 to

4,6,0,5,0,1,0,2,3

, i.e. three copies of

1

and each of

\omega,\dots,\omega^{6}

once, so

\alpha\beta=3+(-1)=2

. Hence

\alpha,\beta

are roots of

x^{2}+x+2=0

. Now

\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta=1-4=-3

and

\alpha^{2}\beta^{2}=(\alpha\beta)^{2}=4

. The required equation is

x^{2}-(-3)x+4

, that is

x^{2}+3x+4=0

. Option (D) is the equation for

\alpha,\beta

with product wrongly taken as 4; option (B) has the sign of the sum reversed; option (A) uses product

-4

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