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lf x+iy= {3}{2++i} then the value of (x-3)(x-1)+y^{2}=

Question

lf $x + i y = \frac{3}{2 + cos θ + i sin θ}$ then the value of $(x - 3) (x - 1) + y^{2} =$

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Daniel Discussed

lf

x + i y = \frac{3}{2 + cos θ + i sin θ}

then the value of

(x - 3) (x - 1) + y^{2} =

8 mins ago

Discuss this question LIVE

8 mins ago

Text solutionVerified

Given

x + i y = \frac{3}{2 + cos θ + i sin θ}

By multiplying the numerator and denominator with

2 + cos θ - i sin θ

We get

x + i y = \frac{3 ( 2 + cos θ - i sin θ )}{( 2 + cos θ ) ^{2} + sin ^{2} θ} = \frac{3 ( 2 + cos θ )}{5 + 4 cos θ} + i \frac{3 ( - sin θ )}{5 + 4 cos θ}

\Rightarrow x = \frac{3 ( 2 + cos θ )}{5 + 4 cos θ}, y = \frac{3 ( - sin θ )}{5 + 4 cos θ}

(x - 3) (x - 1) + y^{2} = (\frac{- 9 ( 1 + cos θ )}{5 + 4 cos θ}) (\frac{1 - cos θ}{5 + 4 cos θ}) + (\frac{3 ( - sin θ )}{5 + 4 cos θ})^{2} = \frac{- 9 + 9 cos ^{2} θ + 9 sin ^{2} θ}{( 5 + 4 cos θ ) ^{2}} = 0

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Question Text	lf $x + i y = \frac{3}{2 + cos θ + i sin θ}$ then the value of $(x - 3) (x - 1) + y^{2} =$
Answer Type	Text solution:1
Upvotes	150