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Question

Mathematics Question on complex numbers

The real part of (1−cos θ+i sin θ)−1( {1 - cos \, \theta + i \, sin \, \theta} )^{-1} is

A

12\frac{1}{2}

B

11+cos⁡ θ\frac{1}{1 + \cos \, \theta}

C

tan⁡θ2\tan \frac{\theta}{2}

D

cot⁡θ2\cot \frac{\theta}{2}

Answer

12\frac{1}{2}

Explanation

Solution

We have, (1−cos⁡ θ+i sin⁡ θ)−1(1- \cos \,\theta + i \,\sin \,\theta)^{-1} =11−cos⁡ θ+i sin⁡ θ=\frac{1}{1-\cos\, \theta+i\, \sin\, \theta} =1−cos⁡ θ−i sin⁡ θ(1−cos⁡ θ)2+sin⁡2θ=\frac{1-\cos\, \theta-i \,\sin \,\theta}{(1-\cos \,\theta)^{2}+\sin ^{2} \theta} =1−cos⁡ θ−i sin⁡ θ1−2 cos⁡ θ+cos⁡2θ+sin⁡2θ=\frac{1-\cos\, \theta-i \,\sin\, \theta}{1-2 \,\cos \,\theta+\cos ^{2} \theta+\sin ^{2} \theta} =1−cos⁡ θ−i sin θ2−2 cos⁡ θ=\frac{1-\cos\, \theta-i \\\,sin\, \theta}{2-2 \,\cos \,\theta} =1−cos⁡ θ2(1−cos⁡ θ)−i sin⁡ θ2(1−cos⁡ θ)=\frac{1-\cos\, \theta}{2(1-\cos \,\theta)}-\frac{i\, \sin\, \theta}{2(1-\cos \,\theta)} =12−i sin⁡ θ2(1−cos⁡ θ)=\frac{1}{2}-\frac{i \,\sin\, \theta}{2(1-\cos\, \theta)} Hence, the real part is 12\frac{1}{2}.