If (cos theta + isin theta )(cos 2theta + isi | vedclass

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(C) Given the product of complex numbers in polar form: $(\cos 	heta + i\sin 	heta )(\cos 2	heta + i\sin 2	heta ) \dots (\cos n	heta + i\sin n	h

Vedclass · Accepted Answer

$\frac{4m\pi}{n(n + 1)}$

Vedclass · Answer

(C) Given the product of complex numbers in polar form: $(\cos 	heta + i\sin 	heta )(\cos 2	heta + i\sin 2	heta ) \dots (\cos n	heta + i\sin n	heta ) = 1$. Using the property $(\cos \alpha + i\sin \alpha)(\cos \beta + i\sin \beta) = \cos(\alpha + \beta) + i\sin(\alpha + \beta)$,we get: $\cos(	heta + 2	heta + 3	heta + \dots + n	heta) + i\sin(	heta + 2	heta + 3	heta + \dots + n	heta) = 1$. The sum of the first $n$ natural numbers is $\frac{n(n + 1)}{2}$,so the expression becomes: $\cos\left(\frac{n(n + 1)}{2}	hetaight) + i\sin\left(\frac{n(n + 1)}{2}	hetaight) = 1$. For this to equal $1$ (which is $\cos(2m\pi) + i\sin(2m\pi)$),the argument must be an integer multiple of $2\pi$: $\frac{n(n + 1)}{2}	heta = 2m\pi$,where $m \in \mathbb{Z}$. Solving for $	heta$: $	heta = \frac{4m\pi}{n(n + 1)}$.

If $(\cos \theta + i\sin \theta )(\cos 2\theta + i\sin 2\theta ) \dots (\cos n\theta + i\sin n\theta ) = 1$,then the value of $\theta$ is

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