Test Index

UPSC NDA II 2023 Mathematics Solved Paper

Question Numbers: 33-35

Consider the following for the next items that follow:

Consider equation-I : z3 + 2z2 + 2z + 1 = 0 and equation-II : z1985 + z100 + 1 = 0.

Question : 34 of 120

Marks: +1, -0

Which one of the following is a root of equation-II?

Solution:

Concept:

The cube roots of unity

1

\omega

\omega^2

satisfy

\omega^3=1

and

1+\omega+\omega^2=0

Explanation:

The given equation is

z^{1985}+z^{100}+1=0

Test each option by substitution.

For

z=-1

(-1)^{1985}+(-1)^{100}+1 = -1+1+1=1\neq0

. So

-1

is not a root.

For

z=-\omega

(-\omega)^{1985}+(-\omega)^{100}+1

. Since

1985

is odd,

(-\omega)^{1985}=-\omega^{1985}=-\omega^{3\cdot661+2}=-\omega^2

;

(-\omega)^{100}=(-\omega)^{4\cdot25}=((-\omega)^4)^{25}=(\omega^4)^{25}=(\omega)^{100}= \omega^{3\cdot33+1}=\omega

. Sum

=-\omega^2+\omega+1

. Using

1+\omega+\omega^2=0

, we have

\omega+1=-\omega^2

, so

-\omega^2+\omega+1 = -\omega^2+(-\omega^2)=-2\omega^2\neq0

. So

-\omega

is not a root.

For

z=-\omega^2

(-\omega^2)^{1985}+(-\omega^2)^{100}+1 = -\omega^{3970}+\omega^{200}+1

\omega^{3970}=\omega^{3\cdot1323+1}=\omega

\omega^{200}=\omega^{3\cdot66+2}=\omega^2

. So

-\omega+\omega^2+1

. Using

1=-\omega-\omega^2

, we get

-\omega+\omega^2+(-\omega-\omega^2)=-2\omega\neq0

. So

-\omega^2

is not a root.

For

z=\omega

\omega^{1985}+\omega^{100}+1 = \omega^{3\cdot661+2}+\omega^{3\cdot33+1}+1 = \omega^2+\omega+1=0

. Hence

\omega

satisfies the equation.

Answer:

\omega

(Option D)

Go to Question:

Prev Question

Next Question