Let $z_{k}=\cos \left(\frac{2 k \pi}{10}\right)+i \sin \left(\frac{2 k \pi}{10}\right) ; k=1,2, \ldots, 9$. | List I | | List II | | --- | --- | --- | | P. | For each $z_{k}$ there exists a $z_{j}$ such that $z_{k} \cdot z_{j}=1$ | 1. True | | Q. | There 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. | 2. False | | R. | $\frac{\left|1-z_{1}\right|\left|1-z_{2}\right| \cdots\left|1-z_{9}\right|}{10}$ equals | 3. 1 | | S. | $1-\sum_{k=1}^{9} \cos \left(\frac{2 k \pi}{10}\right)$ equals | 4. 2 | P Q R S

Complex Numbers — JEE Advanced · 2014 · 25 May 2014 · Paper 2 Question with Solution

Let

z_{k} = cos (\frac{2 k π}{10}) + i sin (\frac{2 k π}{10}); k = 1, 2, \dots, 9

List I		List II
P.	For each $z_{k}$ there exists a $z_{j}$ such that $z_{k} \cdot z_{j} = 1$	1. True
Q.	There exists a $k \in {1, 2, \dots, 9}$ such that $z_{1} \cdot z = z_{k}$ has no solution $z$ in the set of complex numbers.	2. False
R.	$\frac{∣ 1 - z _{1} ∣ ∣ 1 - z _{2} ∣ \dots ∣ 1 - z _{9} ∣}{10}$ equals	3. 1
S.	$1 - \sum_{k = 1}^{9} cos (\frac{2 k π}{10})$ equals	4. 2

P Q R S

Answer: (C)

P → 1 (True): For each

z_{k} = e^{i \frac{2 k π}{10}}

, there exists

z_{j} = e^{i \frac{2 ( 10 - k ) π}{10}}

such that

z_{k} \cdot z_{j} = 1

.

Q → 2 (False): The equation

z_{1} \cdot z = z_{k}

gives

z = \frac{z _{k}}{z _{1}}

, which always has a solution in complex numbers.

R → 3 (equals 1): The values

z_{1}, z_{2}, \dots, z_{9}

along with 1 are the 10th roots of unity. They satisfy

z^{10} - 1 = (z - 1) (z - z_{1}) \dots (z - z_{9})

. Dividing by

z - 1

z^{9} + z^{8} + \dots + 1 = (z - z_{1}) \dots (z - z_{9})

. Setting

z = 1

10 = ∣1 - z_{1} ∣ \dots ∣1 - z_{9} ∣

, so the expression equals 1.

S → 4 (equals 2): The sum of all 10th roots of unity is 0:

1 + z_{1} + \dots + z_{9} = 0

. Taking real parts:

1 + \sum_{k = 1}^{9} cos (\frac{2 k π}{10}) = 0

, so

1 - \sum_{k = 1}^{9} cos (\frac{2 k π}{10}) = 2

.

Answer: P→1, Q→2, R→3, S→4, which is 1234.

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option