SATHEE: Complex Numbers 5 Question 12

Complex Numbers 5 Question 12

13. Let $z_{k} = \cos \frac{2 k π}{10} + i \sin \frac{2 k π}{10}; k = 1, 2, \dots 9$ .

Column I	Column II
P.	For each $z_{k}$ , there exists a $z_{j}$ such that $z_{k} \cdot z_{j} = 1$	(i) 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	(ii) False
R. $\frac{\| 1 - z_{1} \| \| 1 - z_{2} \| \dots \| 1 - z_{9} \|}{10}$ equal	(iii) 1
S. $1 - \sum_{k = 1}^{9} \cos \frac{2 k π}{10}$ equals	(iv) 2

Codes

	P	Q	R	S
(a)	(i)	(ii)	(iv)	(iii)
(b)	(ii)	(i)	(iii)	(iv)
(c)	(i)	(ii)	(iii)	(iv)
(d)	(ii)	(i)	(iv)	(iii)

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Answer:

Correct Answer: 13. (c)

Solution:

(P) PLAN $e^{i θ} \cdot e^{i α} = e^{i (θ + α)}$

Given $z_{k} = e^{i \frac{2 k π}{10}} \Rightarrow z_{k} \cdot z_{j} = e^{i \frac{2 π}{10} (k + j)}$

$z_{k}$ is 10 th root of unity.

$\Rightarrow {\bar{z}}_{k}$ will also be 10th root of unity.

Taking, $z_{j}$ as ${\bar{z}}_{k}$ , we have $z_{k} \cdot z_{j} = 1$ (True)

$(Q) PLAN \begin{aligned} \frac{e^{i θ}}{e^{i α}} & = e^{i (θ - α)} \\ z & = z_{k} / z_{1} = e^{i \frac{2 k π}{10} - \frac{2 π}{10}} = e^{i \frac{π}{5} (k - 1)} \end{aligned}$

For $k = 2; z = e^{i \frac{π}{5}}$ which is in the given set (False)

(R) PLAN

(i) $1 - \cos 2 θ = 2 \sin^{2} θ$

(ii) $\sin 2 θ = 2 \sin θ \cos θ$ and

$(i) \cos 36^{\circ} = \frac{\sqrt{5} - 1}{4}$

(ii) $\cos 108^{\circ} = \frac{\sqrt{5} + 1}{4} \frac{| 1 - z_{1} | | 1 - z_{2} | \dots | 1 - z_{9} |}{10}$

NOTE $| 1 - z_{k} | = 1 - \cos \frac{2 π k}{10} - i \sin \frac{2 π k}{10}$

$= 2 \sin \frac{π k}{10} \sin \frac{π k}{10} - i \cos \frac{π k}{10} = 2 | \sin \frac{π k}{10} |$

Now, required product is

$\begin{aligned} \frac{2^{9} \sin \frac{π}{10} \cdot \sin \frac{2 π}{10} \cdot \sin \frac{3 π}{10} \dots \sin \frac{8 π}{10} \cdot \sin \frac{9 π}{10}}{10} \\ = \frac{2^{9} \sin \frac{π}{10} \sin \frac{2 π}{10} \sin \frac{3 π}{10} \sin \frac{4 π}{10}^{2} \sin \frac{5 π}{10}}{10} \\ = \frac{2^{9} \sin \frac{π}{10} \cos \frac{π}{10} \cdot \sin \frac{2 π}{10} \cos \frac{2 π}{10}^{2} \cdot 1}{10} \end{aligned}$

$\begin{aligned} = \frac{2^{9} \frac{1}{2} \sin \frac{π}{5} \cdot \frac{1}{2} \sin \frac{2 π}{5}^{2}}{10} \\ = \frac{2^{5} {(\sin 36^{\circ} \cdot \sin 72^{\circ})}^{2}}{10} \\ = \frac{2^{5}}{2^{2} \times 10} {(2 \sin 36^{\circ} \sin 72^{\circ})}^{2} \\ = \frac{2^{2}}{5} {(\cos 36^{\circ} - \cos 108^{\circ})}^{2} \\ = \frac{2^{2}}{5} \frac{\sqrt{5} - 1}{4} + \frac{\sqrt{5} + 1}{4} = \frac{2^{2}}{5} \cdot \frac{5}{4} = 1 \end{aligned}$

(S) Sum of $n$ th roots of unity $= 0$

$\begin{aligned} 1 + α + α^{2} + α^{3} + \dots + α^{9} & = 0 \\ 1 + \sum_{k = 1}^{9} α^{k} & = 0 \\ 1 + \sum_{k = 1}^{9} \cos \frac{2 k π}{10} + i \sin \frac{2 k π}{10} & = 0 \\ 1 + \sum_{k = 1}^{9} \cos \frac{2 k π}{10} & = 0 \\ 1 - \sum_{k = 1}^{9} \cos \frac{2 k π}{10} & = 2 \end{aligned}$

$1 ω ω^{2}$

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अगला

Complex Numbers 5 Question 12

13. Let $z_{k} = \cos \frac{2 k π}{10} + i \sin \frac{2 k π}{10}; k = 1, 2, \dots 9$ .

Codes

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Answer:

(R) PLAN

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

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परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

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Complex Numbers 5 Question 12

13. Let zk=cos⁡2kπ10+isin⁡2kπ10;k=1,2,…9.

Codes

Fill in the Blanks

Answer:

(R) PLAN

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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13. Let $z_{k} = \cos \frac{2 k π}{10} + i \sin \frac{2 k π}{10}; k = 1, 2, \dots 9$ .