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VITEEE 2014 Paper

Section: Mathematics

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Question : 104 of 120

Marks: +1, -0

If x + iy =

\frac{3}{2+\cos\theta+i\sin\theta}

x^2+y^2

is equal to

3x - 4
4x - 3
4x + 3
None of these

Solution:

x + iy

=

\frac{3}{2+\cos\theta+i\sin\theta}

=

\frac{3(2+\cos\theta-i\sin\theta)}{(2+\cos\theta)^2+\sin^2\theta}

=

\frac{6+3\cos\theta-3i\sin\theta}{4+\cos^2\theta+4\cos\theta+\sin^2\theta}

=

\frac{6+3\cos\theta-3i\sin\theta}{5+4\cos\theta}

=

\left(\frac{6+3\cos\theta}{5+4\cos\theta}\right)

+

\left(\frac{-3\sin\theta}{5+4\cos\theta}\right)

On equating real and imaginary parts, we get

x =

\frac{3(2+\cos\theta)}{5+4\cos\theta}

And y =

\frac{-3\sin\theta}{5+4\cos\theta}

∴

x^2+y^2

=

\frac{9[4+\cos^2\theta+4\cos\theta+\sin^2\theta]}{(5+4\cos\theta)^2}

=

\frac{9}{5+4\cos\theta}

= 4

\left(\frac{6+3\cos\theta}{5+4\cos^2\theta}\right)

- 3 = 4x - 3

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