Question

Let ${z_k} = \cos \left( {\dfrac{{2k\pi }}{{10}}} \right) + i\sin \left( {\dfrac{{2k\pi }}{{10}}} \right); k = 1,2,......9.$

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 exist a $k \in \left( {1,2,......,9} \right)$ such that ${z_1} \cdot z = {z_k}$ has no solution $z$ in the set of complex numbers	(2) False
(R) $\dfrac{{\left| {1 - {z_1}} \right|\left| {1 - {z_2}} \right|......\left| {1 - {z_9}} \right|}}{{10}}$ is equals to	(3) 1
(S) $1 - \sum\nolimits_{k = 1}^9 {\cos \left( {\dfrac{{2k\pi }}{{100}}} \right)} $	(4) 2

Which of the following is the correct match?
A. P-1, Q-2, R-4, S-3
B. P-2, Q-1, R-3, S-4
C. P-1, Q-2, R-3, S-4
D. P-2, Q-1, R-4, S-3

Answer 1

Hint: A number of the form $z = a + ib$ , where $a{\text{ and }}b$ are real numbers, is called a complex number, $a$ is called the real part, and $b$ called the imaginary part.
The conjugate of the complex number $z = a + ib$ , denoted by $\bar z$,
$\bar z = a - ib$
If $n$ is a positive integer, then $'x'$ is said to be an ${n^{th}}$ root of unity if it satisfies the equation ${x^n} = 1$ .
For example: For $n = 3$ , $'x'$ is said to be an ${3^{rd}}$ or cubic root of unity of equation ${x^3} = 1$ .
The complex expression ${x_k} = \cos \left( {\dfrac{{2k\pi }}{n}} \right) + i\sin \left( {\dfrac{{2k\pi }}{n}} \right);n = 1,2,......,\left( {n - 1} \right)$ is the ${n^{th}}$ root of unity.
You can also derive the ${n^{th}}$ root of unity but it is advisable to remember it.

Complete step-by-step answer:
Given that ${z_k} = \cos \left( {\dfrac{{2k\pi }}{{10}}} \right) + i\sin \left( {\dfrac{{2k\pi }}{{10}}} \right);k = 1,2,......,9.$
${z_k}$ is a complex number. For every complex number, there exists its conjugate $\overline {{z_k}} $
For $k = 1$, ${z_1} = \cos \left( {\dfrac{\pi }{5}} \right) + i\sin \left( {\dfrac{\pi }{5}} \right)$
Then its conjugate, $\overline {{z_1}} = \cos \left( {\dfrac{\pi }{5}} \right) - i\sin \left( {\dfrac{\pi }{5}} \right)$
Let's find ${z_1} \cdot \overline {{z_1}} $
${z_1} \cdot \overline {{z_1}} = \left[ {\cos \left( {\dfrac{\pi }{5}} \right) + i\sin \left( {\dfrac{\pi }{5}} \right)} \right]\left[ {\cos \left( {\dfrac{\pi }{5}} \right) - i\sin \left( {\dfrac{\pi }{5}} \right)} \right]$
Using identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$
On comparing with the above identity:
 $a = \cos \left( {\dfrac{\pi }{5}} \right)$ , $b = i\sin \left( {\dfrac{\pi }{5}} \right)$
$
  {z_1} \cdot \overline {{z_1}} = {\left( {\cos \left( {\dfrac{\pi }{5}} \right)} \right)^2} - {\left( {i\sin \left( {\dfrac{\pi }{5}} \right)} \right)^2} \\
   \Rightarrow {\left( {\cos \left( {\dfrac{\pi }{5}} \right)} \right)^2} - {i^2}{\left( {\sin \left( {\dfrac{\pi }{5}} \right)} \right)^2} \\
 $
We know, ${i^2} = - 1$ , substituting it.
${z_1} \cdot \overline {{z_1}} = {\left( {\cos \left( {\dfrac{\pi }{5}} \right)} \right)^2} + {\left( {i\sin \left( {\dfrac{\pi }{5}} \right)} \right)^2}$
Using trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$
${z_1} \cdot \overline {{z_1}} = 1$
This result is true for every $k \in \left( {1,2,......,9} \right)$
 so that \[{z_k} \cdot \overline {{z_k}} = 1\]
$ \Rightarrow \overline {{z_k}} = \cos \left( {\dfrac{{2k\pi }}{{10}}} \right) - i\sin \left( {\dfrac{{2k\pi }}{{10}}} \right);k = 1,2,......,9.$
Take ${z_j} = \overline {{z_k}} $, then the given statement:
For each ${z_k}$ there exists a ${z_j}$ such that ${z_k} \cdot {z_j} = 1$ is true.

Solving (Q)
The given equation ${z_1} \cdot z = {z_k}$
Multiplying both sides by the conjugate of ${z_1}$ . i.e. $\overline {{z_1}} $
$\overline {{z_1}} {z_1} \cdot z = \overline {{z_1}} \cdot {z_k}$
We know, ${z_1} \cdot \overline {{z_1}} = 1$
$
   \Rightarrow 1 \cdot z = \overline {{z_1}} \cdot {z_k} \\
   \Rightarrow z = \overline {{z_1}} \cdot {z_k} \\
 $
The term $\overline {{z_1}} \cdot {z_k}$ is a multiplication of two complex numbers $\overline {{z_1}} {\text{ and }}{z_k}$ , it will be either real or complex number. In other words, the product $\overline {{z_1}} \cdot {z_k}$ exists.
Hence, the value of $z$ in equation ${z_1} \cdot z = {z_k}$ exists.
Therefore, the given statement:
There exist a $k \in \left( {1,2,......,9} \right)$ such that ${z_1} \cdot z = {z_k}$ has no solution $z$ in the set of complex numbers is false

Solving (R):
On comparing with ${x_k} = \cos \left( {\dfrac{{2k\pi }}{n}} \right) + i\sin \left( {\dfrac{{2k\pi }}{n}} \right);k = 1,2,......,\left( {k - 1} \right)$
The given expression ${z_k} = \cos \left( {\dfrac{{2k\pi }}{{10}}} \right) + i\sin \left( {\dfrac{{2k\pi }}{{10}}} \right);k = 1,2,......,9.$
Value of n = 10
So, ${z_k}$ is the tenth root of unity.
The expansion of the equation ${x^{10}} - 1 = 0$ gives the roots.
${x^{10}} - 1 = \left( {x - 1} \right)\left( {x - {z_1}} \right)\left( {x - {z_2}} \right)......\left( {x - {z_9}} \right)$
$\dfrac{{{x^{10}} - 1}}{{\left( {x - 1} \right)}} = \left( {x - {z_1}} \right)\left( {x - {z_2}} \right)......\left( {x - {z_9}} \right)$
The expression $\dfrac{{{x^{10}} - 1}}{{\left( {x - 1} \right)}}$ is the sum of Geometric Progression, G.P. $1,{x^1},{x^2},{x^3},......,{x^9}$
$\dfrac{{{x^{10}} - 1}}{{\left( {x - 1} \right)}}$ $ = 1 + {x^1} + {x^2} + {x^3} + ...... + {x^9}$
$ \Rightarrow 1 + x + {x^2} + ...... + {x^9} = \left( {x - {z_1}} \right)\left( {x - {z_2}} \right)......\left( {x - {z_9}} \right)$
Taking modulus on both sides
$ \Rightarrow \left| {1 + x + {x^2} + ...... + {x^9}} \right| = \left| {\left( {x - {z_1}} \right)\left( {x - {z_2}} \right)......\left( {x - {z_9}} \right)} \right|$
Put $x = 1$
$ \Rightarrow \left| {1 + 1 + 1 + ...... + 1} \right| = \left| {\left( {1 - {z_1}} \right)} \right|\left| {\left( {1 - {z_2}} \right)} \right|......\left| {\left( {1 - {z_9}} \right)} \right|$
$ \Rightarrow \left| {\left( {1 - {z_1}} \right)} \right|\left| {\left( {1 - {z_2}} \right)} \right|......\left| {\left( {1 - {z_9}} \right)} \right| = 10$
Given expression $\dfrac{{\left| {1 - {z_1}} \right|\left| {1 - {z_2}} \right|......\left| {1 - {z_9}} \right|}}{{10}}$
$ \Rightarrow \dfrac{{10}}{{10}} = 1$

Solving (S):
In the equation ${x^{10}} - 1 = 0$
Sum of its roots = 0
Given ${z_k}$ is the tenth root of unity.
$ \Rightarrow \sum\limits_{k = 0}^9 {{z_k}} = 0$
Given that ${z_k} = \cos \left( {\dfrac{{2k\pi }}{{10}}} \right) + i\sin \left( {\dfrac{{2k\pi }}{{10}}} \right);k = 1,2,......,9.$
$\sum\limits_{k = 0}^9 {\left[ {\cos \left( {\dfrac{{2k\pi }}{{10}}} \right) + i\sin \left( {\dfrac{{2k\pi }}{{10}}} \right)} \right]} = 0$
$ \Rightarrow \sum\limits_{k = 0}^9 {\cos \left( {\dfrac{{2k\pi }}{{10}}} \right)} + i\sum\limits_{k = 0}^9 {\sin \left( {\dfrac{{2k\pi }}{{10}}} \right)} = 0$
The term \[\sum\limits_{k = 0}^9 {\cos \left( {\dfrac{{2k\pi }}{{10}}} \right)} = \cos \left( {\dfrac{{2\left( 0 \right)\pi }}{{10}}} \right) + \cos \left( {\dfrac{{2\left( 1 \right)\pi }}{{10}}} \right) + \cos \left( {\dfrac{{2\left( 2 \right)\pi }}{{10}}} \right) + ...\] will be some real number, and similarly, the term $\sum\limits_{k = 0}^9 {\sin \left( {\dfrac{{2k\pi }}{{10}}} \right)} $ will also be some real number.
For the complex number $a + ib = 0$ , $a = 0,b = 0$.
That is why $\sum\limits_{k = 0}^9 {\cos \left( {\dfrac{{2k\pi }}{{10}}} \right)} = 0$ and $\sum\limits_{k = 0}^9 {\sin \left( {\dfrac{{2k\pi }}{{10}}} \right)} = 0$
$ \Rightarrow \cos \left( {\dfrac{{2\left( 0 \right)\pi }}{{10}}} \right) + \sum\limits_{k = 1}^9 {\cos \left( {\dfrac{{2k\pi }}{{10}}} \right)} = 0$
$ \Rightarrow \cos \left( 0 \right) + \sum\limits_{k = 1}^9 {\cos \left( {\dfrac{{2k\pi }}{{10}}} \right)} = 0$
We know, $\cos \left( 0 \right) = 1$
$ \Rightarrow \sum\limits_{k = 1}^9 {\cos \left( {\dfrac{{2k\pi }}{{10}}} \right)} = 1$
The given expression in (S):
$1 - \sum\nolimits_{k = 1}^9 {\cos \left( {\dfrac{{2k\pi }}{{100}}} \right)} $
$ \Rightarrow 1 - \left( { - 1} \right) = 2$

So, the correct answer is “Option C”.

Note: In the trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$ make sure that the angle of sine and cosine is the same.
The value of unknown (or variable) is an equation that is known as its solution.
Notice that for $\sum\limits_{k = 0}^9 {\sin \left( {\dfrac{{2k\pi }}{{10}}} \right)} = 0$
$\sin \left( 0 \right) + \sum\limits_{k = 1}^9 {\sin \left( {\dfrac{{2k\pi }}{{10}}} \right)} = 0$
We know $\sin 0 = 0$
Thus, the summation from $k = 1{\text{ to 9}}$, $\sum\limits_{k = 1}^9 {\sin \left( {\dfrac{{2k\pi }}{{10}}} \right)} $ is still 0.