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Question: If \(x + iy = \frac{3}{\cos\theta + i\sin\theta + 2}\), then \(4x - x^{2} - y^{2}\)reduces to...

If x+iy=3cos⁡θ+isin⁡θ+2x + iy = \frac{3}{\cos\theta + i\sin\theta + 2}, then 4x−x2−y24x - x^{2} - y^{2}reduces to

A

2

B

3

C

4

D

5

Answer

3

Explanation

Solution

Sol. x−iyx2+y2=1x+iy=13[cos⁡θ+2+isin⁡θ]\frac{x - iy}{x^{2} + y^{2}} = \frac{1}{x + iy} = \frac{1}{3}\left\lbrack \cos\theta + 2 + i\sin\theta \right\rbrack

⇒xx2+y2=13(cos⁡θ+2),−yx2+y2=13sin⁡θ\frac{x}{x^{2} + y^{2}} = \frac{1}{3}\left( \cos\theta + 2 \right),\frac{- y}{x^{2} + y^{2}} = \frac{1}{3}\sin\theta

⇒(xx2+y2−23)2+(−yx2+y2)2=19\left( \frac{x}{x^{2} + y^{2}} - \frac{2}{3} \right)^{2} + \left( \frac{- y}{x^{2} + y^{2}} \right)^{2} = \frac{1}{9}

⇒x2+y2(x2+y2)2−4x3(x2+y2)2+13=0\frac{x^{2} + y^{2}}{\left( x^{2} + y^{2} \right)^{2}} - \frac{4x}{3\left( x^{2} + y^{2} \right)^{2}} + \frac{1}{3} = 0

⇒ 1x2+y2(3−4x)+1=0\frac{1}{x^{2} + y^{2}}(3 - 4x) + 1 = 0

⇒ 4x - x2 - y2 = 3