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Question: \((\cos\theta + i\sin\theta)(\cos 2\theta + i\sin 2\theta)........\) upto \(a = - (\sqrt{2})i\) term...

(cos⁡θ+isin⁡θ)(cos⁡2θ+isin⁡2θ)........(\cos\theta + i\sin\theta)(\cos 2\theta + i\sin 2\theta)........ upto a=−(2)ia = - (\sqrt{2})i terms

A

a=1−ia = 1 - i

B

a=1+ia = 1 + i

C

1

D

a=2ia = \sqrt{2i}

Answer

a=2ia = \sqrt{2i}

Explanation

Solution

Given expression isz1z_{1}upto z2z_{2} terms.

Obviously number of terms is odd, so expression has the value –1.