SATHEE: Complex Numbers 4 Question 1

Complex Numbers 4 Question 1

1. Let $z = \frac{\sqrt{3}}{2} + {\frac{i}{2}}^{5} + \frac{\sqrt{3}}{2} - {\frac{i}{2}}^{5}$ . If $R (z)$ and $I (z)$ respectively denote the real and imaginary parts of $z$ , then

(2019 Main, 10 Jan II)

(a) $R (z) > 0$ and $I (z) > 0$

(b) $I (z) = 0$

(d) $R (z) = - 3$

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Answer:

Correct Answer: 1. (b)

Solution:

Given, $z = \frac{\sqrt{3}}{2} + {\frac{i}{2}}^{5} + \frac{\sqrt{3}}{2} - {\frac{i}{2}}^{5}$

$∵$ Euler’s form of

$\begin{aligned} \begin{matrix} \frac{\sqrt{3}}{2} + \frac{i}{2} = \cos \frac{π}{6} + i \sin \frac{π}{6} = e^{i (π / 6)} \\ and \frac{\sqrt{3}}{2} - \frac{i}{2} = \cos \frac{- π}{6} + i \sin - \frac{π}{6} = e^{- i π / 6} \\ So, z = {(e^{i π / 6})}^{5} + {(e^{- i π / 6})}^{5} = e^{i \frac{5 π}{6}} + e^{- i \frac{5 π}{6}} \\ = \cos \frac{5 π}{6} + i \sin \frac{5 π}{6} + \cos \frac{5 π}{6} - i \sin \frac{5 π}{6} \\ = 2 \cos \frac{5 π}{6} \end{matrix} [∵ e^{i θ} = \cos θ + i \sin θ] \end{aligned}$

$∴ I (z) = 0$ and $R (z) = - 2 \cos \frac{π}{6} = - \sqrt{3} < 0$

$∵ \cos \frac{5 π}{6} = \cos π - \frac{π}{6} = - \cos \frac{π}{6}$

Complex Numbers 4 Question 1

1. Let $z = \frac{\sqrt{3}}{2} + {\frac{i}{2}}^{5} + \frac{\sqrt{3}}{2} - {\frac{i}{2}}^{5}$ . If $R (z)$ and $I (z)$ respectively denote the real and imaginary parts of $z$ , then

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Complex Numbers 4 Question 1

1. Let z=32+i25+32−i25. If R(z) and I(z) respectively denote the real and imaginary parts of z, then

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1. Let $z = \frac{\sqrt{3}}{2} + {\frac{i}{2}}^{5} + \frac{\sqrt{3}}{2} - {\frac{i}{2}}^{5}$ . If $R (z)$ and $I (z)$ respectively denote the real and imaginary parts of $z$ , then