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Question: If \(x + iy\), where \(a = \cos\theta + i\sin\theta,\) are real, then \(\frac{1 + a}{1 - a} =\)...

If x+iyx + iy, where a=cos⁡θ+isin⁡θ,a = \cos\theta + i\sin\theta, are real, then 1+a1−a=\frac{1 + a}{1 - a} =

A

1

B

cot⁡θ\cot\theta

C

cot⁡θ2\cot\frac{\theta}{2}

D

icot⁡θ2i\cot\frac{\theta}{2}

Answer

1

Explanation

Solution

z=(2+i)23+iz = \frac{(2 + i)^{2}}{3 + i} .....(i)

∴=3+4i3+i×3−i3−i= \frac{3 + 4i}{3 + i} \times \frac{3 - i}{3 - i} .....(ii)

Multiplying (i) and (ii), we get

=1310+i910\mathbf{=}\frac{\mathbf{13}}{\mathbf{10}}\mathbf{+ i}\frac{\mathbf{9}}{\mathbf{10}} =1310−i910= \frac{13}{10} - i\frac{9}{10}.