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Question

Mathematics Question on complex numbers

If a=eiθa = e^{i \theta} , then 1+a1−a\frac{1 + a}{1-a} is equal to

A

cot⁡θ2\cot \frac{\theta}{2}

B

tan⁡θ\tan \theta

C

i cot⁡θ2i \, \cot \frac{\theta}{2}

D

i tan⁡θ2i \, \tan \frac{\theta}{2}

Answer

i cot⁡θ2i \, \cot \frac{\theta}{2}

Explanation

Solution

We have,
a=eiθa=e^{i \theta}
=cos⁡θ+i sin⁡ θ=\cos \theta+i \,\sin \,\theta
Now, 1+a1−a=1+(cos⁡ θ+isin⁡ θ)1−(cos⁡ θ+i sin⁡ θ)\frac{1+a}{1-a}=\frac{1+(\cos\, \theta+i \sin \,\theta)}{1-(\cos \,\theta+i \,\sin\,\theta)}
=(1+cos⁡ θ)+i sin⁡ θ(1−cos⁡ θ)−i sin⁡ θ=\frac{(1+\cos\, \theta)+i\, \sin\, \theta}{(1-\cos \,\theta)-i \,\sin\, \theta}
=2 cos⁡2θ2+i 2 sin⁡θ2 cos⁡θ22 sin⁡2θ2−i 2 sin⁡θ2 cos⁡θ2=\frac{2 \,\cos ^{2} \frac{\theta}{2}+i \,2 \,\sin \frac{\theta}{2} \,\cos \frac{\theta}{2}}{2 \,\sin ^{2} \frac{\theta}{2}-i\, 2 \,\sin \frac{\theta}{2} \,\cos \frac{\theta}{2}}
=2 cos⁡θ2[cos⁡θ2+i sin⁡θ2]2 sin⁡θ2[sin⁡θ2−icos⁡θ2]=\frac{2\, \cos \frac{\theta}{2}\left[\cos \frac{\theta}{2}+i \,\sin \frac{\theta}{2}\right]}{2 \,\sin \frac{\theta}{2}\left[\sin \frac{\theta}{2}-i \cos \frac{\theta}{2}\right]}
=cot⁡θ2[cos⁡θ2+i sin⁡θ2]−i [cos⁡θ2+i sin⁡θ2]=\frac{\cot \frac{\theta}{2}\left[\cos \frac{\theta}{2}+i \,\sin \frac{\theta}{2}\right]}{-i\,\left[\cos \frac{\theta}{2}+i\, \sin \frac{\theta}{2}\right]}
cot⁡θ2−i\frac{\cot \frac{\theta}{2}}{-i}
=icot⁡θ2= i \cot \frac{\theta}{2}