Complex NumbershardPYQ 2014 · JEE AdvancedFree

Complex Numbers: Let Match Entry List Correct Entry List (JEE Advanced 2014)

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

Question

Let

z_k = \cos\left(\dfrac{2k\pi}{10}\right) + i\sin\left(\dfrac{2k\pi}{10}\right)

;

k = 1, 2, \ldots, 9

. Match each entry in List-I with the correct entry in List-II.

List-I

PFor each

z_k

, there exists a

z_j

such that

z_k \cdot z_j = 1

QThere exists a

k \in \{1, 2, \ldots, 9\}

such that

z_1 \cdot z = z_k

has no solution

z

in the set of complex numbers.

\dfrac{\left|1 - z_1\right|\left|1 - z_2\right|\cdots\left|1 - z_9\right|}{10}

equals

1 - \displaystyle\sum_{k=1}^{9} \cos\left(\dfrac{2k\pi}{10}\right)

equals

List-II

1 True

2 False

1

2

\to

1, Q

\to

2, R

\to

4, S

\to

\to

2, Q

\to

1, R

\to

3, S

\to

\to

1, Q

\to

2, R

\to

3, S

\to

4correct

\to

2, Q

\to

1, R

\to

4, S

\to

Solution

Step 1: Express

z_k

in exponential form.

z_k = \cos\left(\frac{2k\pi}{10}\right) + i\sin\left(\frac{2k\pi}{10}\right) = e^{\,i\,\frac{2k\pi}{10}}, \quad k = 1, 2, \ldots, 9.

These are the nine non-trivial

10

th roots of unity, that is, all the

10

th roots of unity except

1

. Part (P): Step 1: The condition

z_k \cdot z_j = 1

requires

z_j = \frac{1}{z_k} = e^{-i\,\frac{2k\pi}{10}} = \overline{z_k}.

Step 2: Since

z_k

is a

10

th root of unity, its conjugate satisfies

\overline{z_k} = e^{-i\,\frac{2k\pi}{10}} = e^{\,i\,\frac{2(10-k)\pi}{10}} = z_{10-k},

and for

k \in \{1, 2, \ldots, 9\}

the index

10 - k

also lies in

\{1, 2, \ldots, 9\}

. Hence

z_j = z_{10-k}

is one of the given numbers. Step 3: With this choice,

z_k \cdot z_j = z_k \cdot \overline{z_k} = |z_k|^2 = 1.

Therefore such a

z_j

exists for each

z_k

, and the statement is true. Conclusion: (P)

\to

(1). Part (Q): Step 1: The equation

z_1 \cdot z = z_k

gives

z = \frac{z_k}{z_1}, \quad \text{which is valid since } z_1 = e^{\,i\,\frac{2\pi}{10}} \neq 0.

Step 2: Thus a solution

z \in \mathbb{C}

exists for every

k \in \{1, 2, \ldots, 9\}

. There is no value of

k

for which the equation fails to have a solution. Therefore the statement is false. Conclusion: (Q)

\to

(2). Part (R): Step 1: Since

1, z_1, z_2, \ldots, z_9

are precisely the

10

th roots of unity,

z^{10} - 1 = (z - 1)(z - z_1)(z - z_2)\cdots(z - z_9).

Step 2: Dividing both sides by

(z - 1)

(z - z_1)(z - z_2)\cdots(z - z_9) = \frac{z^{10} - 1}{z - 1} = 1 + z + z^2 + \cdots + z^9.

Step 3: Substituting

z = 1

(1 - z_1)(1 - z_2)\cdots(1 - z_9) = \underbrace{1 + 1 + \cdots + 1}_{10 \text{ terms}} = 10.

Step 4: Taking the modulus of both sides,

|1 - z_1|\,|1 - z_2|\cdots|1 - z_9| = 10.

Therefore

\frac{|1 - z_1|\,|1 - z_2|\cdots|1 - z_9|}{10} = \frac{10}{10} = 1.

Conclusion: (R)

\to

(3). Part (S): Step 1: The sum of all

10

th roots of unity is zero:

1 + z_1 + z_2 + \cdots + z_9 = 0.

Step 2: Equating real parts,

1 + \sum_{k=1}^{9} \cos\left(\frac{2k\pi}{10}\right) = 0 \implies \sum_{k=1}^{9} \cos\left(\frac{2k\pi}{10}\right) = -1.

Step 3: Therefore

1 - \sum_{k=1}^{9} \cos\left(\frac{2k\pi}{10}\right) = 1 - (-1) = 2.

Conclusion: (S)

\to

(4). Combining all parts:

\text{P} \to 1, \quad \text{Q} \to 2, \quad \text{R} \to 3, \quad \text{S} \to 4,

which corresponds to option (C). Correct answer: (C)

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