SATHEE: Complex Numbers 5 Question 8

Complex Numbers 5 Question 8

8. Let $z_{1}$ and $z_{2}$ be $n$ th roots of unity which subtend a right angled at the origin, then $n$ must be of the form (where, $k$ is an integer)

(2001, 1M)

(a) $4 k + 1$

(b) $4 k + 2$

(d) $4 k$

equal to

(a) $1 - i \sqrt{3}$

(b) $- 1 + i \sqrt{3}$

(d) $- i \sqrt{3}$

(1999, 2M)

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Answer:

Correct Answer: 8. (d)

Solution:

Since, $\arg \frac{z_{1}}{z_{2}} = \frac{π}{2}$

$\begin{array}{lll} \Rightarrow & \frac{z_{1}}{z_{2}} & = \cos \frac{π}{2} + i \sin \frac{π}{2} = i \\ ∴ & \frac{z_{1}^{n}}{z_{2}^{n}} & = (i)^{n} \Rightarrow i^{n} = 1 \\ \Rightarrow & n & = 4 k \end{array}$

Alternate Solution

Since,

$\arg \frac{z_{2}}{z_{1}} = \frac{π}{2}$

$∴ \frac{z_{2}}{z_{1}} = | \frac{z_{2}}{z_{1}} | e^{i \frac{π}{2}}$

$\begin{array}{lll} \Rightarrow & \frac{z_{2}}{z_{1}} & = i \\ \Rightarrow & {\frac{z_{2}}{z_{1}}}^{n} & = i^{n} \end{array}$

$[∵ | z_{1} | = | z_{2} | = 1]$

$∴ z_{1}$ and $z_{2}$ are $n$ th roots of unity.

$z_{1}^{n} = z_{2}^{n} = 1$

$\Rightarrow \frac{z_{2}}{z_{1}} = 1$

$\Rightarrow i^{n} = 1$

$\Rightarrow n = 4 k$ , where $k$ is an integer.

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Complex Numbers 5 Question 8

8. Let $z_{1}$ and $z_{2}$ be $n$ th roots of unity which subtend a right angled at the origin, then $n$ must be of the form (where, $k$ is an integer)

Answer:

Alternate Solution

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Complex Numbers 5 Question 8

8. Let z1 and z2 be nth roots of unity which subtend a right angled at the origin, then n must be of the form (where, k is an integer)

Answer:

Alternate Solution

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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8. Let $z_{1}$ and $z_{2}$ be $n$ th roots of unity which subtend a right angled at the origin, then $n$ must be of the form (where, $k$ is an integer)