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Question: If \({Re}\frac{(1 + i)^{2}}{3 - i}\), then\(- 1/5\) equals....

If Re(1+i)23−i{Re}\frac{(1 + i)^{2}}{3 - i}, then−1/5- 1/5 equals.

A

(15, 20)

B

(20, 15)

C

(1−i)x+(1+i)y=1−3i,(1 - i)x + (1 + i)y = 1 - 3i,

D

None of these

Answer

(15, 20)

Explanation

Solution

=sin⁡θ22sin⁡θ2(1+3cos⁡2θ2)=12(1+3cos⁡2θ2)= \frac{\sin\frac{\theta}{2}}{2\sin\frac{\theta}{2}\left( 1 + 3\cos^{2}\frac{\theta}{2} \right)} = \frac{1}{2\left( 1 + 3\cos^{2}\frac{\theta}{2} \right)}⇒ =15+3cos⁡θ= \frac{1}{5 + 3\cos\theta}

Equating real and imaginary parts, we get (x+iy)1/3=a+ib(x + iy)^{1/3} = a + ib and (x+iy)=(a+ib)3(x + iy) = (a + ib)^{3}.