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Seventh Roots of Unity: Evaluate |(a+b) + abi| | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question

P = \cos\dfrac{2\pi}{7} + i\sin\dfrac{2\pi}{7}

a = P + P^{2} + P^{4}

and

b = P^{3} + P^{5} + P^{6}

, then the value of

\left|(a+b) + abi\right|

(where

i = \sqrt{-1}

) is

\sqrt3

\sqrt5

correct

\sqrt6

2

Solution

Step 1: Identify

P

. In Euler form

P = e^{\,i\frac{2\pi}{7}}

, so

P^{7} = e^{\,2\pi i} = 1, \qquad P \ne 1 .

P

is a seventh root of unity other than

1

. Step 2: Find

a + b

. Together,

a

and

b

use every power from

P^{1}

P^{6}

exactly once:

a + b = P + P^{2} + P^{3} + P^{4} + P^{5} + P^{6} .

Step 3: Use the fact that all seven seventh-roots of unity add to zero, i.e.

1 + P + P^{2} + \cdots + P^{6} = 0

. Hence

a + b = -1 .

Step 4: Find

ab

by expanding, reducing every exponent modulo

7

(since

P^{7} = 1

ab = \left(P + P^{2} + P^{4}\right)\left(P^{3} + P^{5} + P^{6}\right)

= \underbrace{P^{4} + P^{6} + P^{7}}_{\text{from } P} + \underbrace{P^{5} + P^{7} + P^{8}}_{\text{from } P^{2}} + \underbrace{P^{7} + P^{9} + P^{10}}_{\text{from } P^{4}} .

Step 5: Reduce:

P^{7} = 1

P^{8} = P

P^{9} = P^{2}

P^{10} = P^{3}

ab = \left(P^{4} + P^{6} + 1\right) + \left(P^{5} + 1 + P\right) + \left(1 + P^{2} + P^{3}\right)

= \left(P + P^{2} + P^{3} + P^{4} + P^{5} + P^{6}\right) + 3 = -1 + 3 = 2 .

Step 6: Substitute into the required expression.

\left|(a+b) + abi\right| = \left|-1 + 2i\right| = \sqrt{(-1)^{2} + 2^{2}} = \sqrt5 .

Answer: (2).

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