(B) Using Euler's formula, $e^{ix} = \cos x + i \sin x$, the given equation becomes:$e^{i\theta} \cdot e^{i2\theta} \cdot e^{i3\theta} \dots e^{in\the

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(B) Using Euler's formula, $e^{ix} = \cos x + i \sin x$, the given equation becomes:$e^{i\theta} \cdot e^{i2\theta} \cdot e^{i3\theta} \dots e^{in\theta} = 1$$e^{i\theta(1 + 2 + 3 + \dots + n)} = 1$Since the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$, we have:$e^{i\frac{n(n+1)}{2}\theta} = 1$Taking the logarithm or comparing with $e^{i2k\pi} = 1$, we get:$\frac{n(n+1)}{2}\theta = 2k\pi$$\theta = \frac{4k\pi}{n(n+1)}$

Class 11 Mathematics 4-1.Complex numbers De Moivre's theorem and Roots of unity

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The general value of the real angle $\theta$, which satisfies the equation $(\cos \theta + i \sin \theta)(\cos 2\theta + i \sin 2\theta) \dots (\cos n\theta + i \sin n\theta) = 1$ is given by (assuming $k$ is an integer):

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A
$\frac{2k\pi}{n+2}$
B
$\frac{4k\pi}{n(n+1)}$
C
$\frac{4k\pi}{n+1}$
D
$\frac{6k\pi}{n(n+1)}$

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4-1.Complex numbers

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De Moivre's theorem and Roots of unity

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The general value of the real angle $\theta$, which satisfies the equation $(\cos \theta + i \sin \theta)(\cos 2\theta + i \sin 2\theta) \dots (\cos n\theta + i \sin n\theta) = 1$ is given by (assuming $k$ is an integer):

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