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Question: If \(\sum_{k = 0}^{100}i^{k} = x + iy\) and \(x\) then value of k equals....

If ∑k=0100ik=x+iy\sum_{k = 0}^{100}i^{k} = x + iy and xx then value of k equals.

A

2

B

4

C

6

D

1

Answer

4

Explanation

Solution

z=1+2i1−iz = \frac{1 + 2i}{1 - i}

⇒z=1+2i1−i×1+i1+i\Rightarrow z = \frac{1 + 2i}{1 - i} \times \frac{1 + i}{1 + i}

⇒ =−12+i32= \frac{- 1}{2} + i\frac{3}{2}

⇒ 11−cos⁡θ+isin⁡θ\frac{1}{1 - \cos\theta + i\sin\theta}

=1(1−cos⁡θ)+isin⁡θ×(1−cos⁡θ)−isin⁡θ(1−cos⁡θ)−isin⁡θ=(1−cos⁡θ)−isin⁡θ(1−cos⁡θ)2+sin⁡2θ= \frac{1}{(1 - \cos\theta) + i\sin\theta} \times \frac{(1 - \cos\theta) - i\sin\theta}{(1 - \cos\theta) - i\sin\theta} = \frac{(1 - \cos\theta) - i\sin\theta}{(1 - \cos\theta)^{2} + \sin^{2}\theta}

=(1−cos⁡θ)−isin⁡θ2(1−cos⁡θ)= \frac{(1 - \cos\theta) - i\sin\theta}{2(1 - \cos\theta)}

=(1−cos⁡θ)2(1−cos⁡θ)−isin⁡θ2(1−cos⁡θ).= \frac{(1 - \cos\theta)}{2(1 - \cos\theta)} - i\frac{\sin\theta}{2(1 - \cos\theta)}. 1−cos⁡θ2(1−cos⁡θ)=12\frac{1 - \cos\theta}{2(1 - \cos\theta)} = \frac{1}{2}.