Question

if,
x+iy=3÷2+cos theta+i sin theta
then prove,
x^2+y^2=4x-3


plz help....

Answer 1

z=(x+iy)
(x+iy)=r(cos theta+i sin theta)
r cos theta=x and ,r sin theta=y
on square we get,
r²=(x²+y²)
9/4=x²+y²

Answer 2

$\huge\color{Black}\boxed{\colorbox{yellow}{❀Answer❀}}$

$x + iy = \frac{3}{2 + cos \theta + isin \theta} \\ rationalizing \: \: the \: \: denominator \\ = \frac{3}{2 + cos \theta + isin \theta} \times \frac{2 + cos \theta - isin \theta}{2 + cos \theta - isin \theta} \\ = \frac{6 + 3cos \theta - 3isin \theta}{( {2 + cos \theta})^{2} - ( {isin \theta})^{2} } \\ = \frac{6 + 3cos \theta - 3isin \theta}{4 + 4cos \theta + {cos}^{2} \theta + {sin}^{2} \theta} \\ = \frac{6 + 3cos \theta - 3isin \theta}{5 + 4cos \theta} \\ = \frac{6 + 3cos \theta}{5 + 4cos \theta} + \frac{ - 3isin \theta}{5 + 4cos \theta} \\ \therefore \: \: \: \: \: \: x = \frac{6 + 3cos \theta}{5 + 4cos \theta} \: \: and \: \: y = \frac{ - 3sin \theta}{5 + 4cos \theta} \\ now \: \: \: {x}^{2} + {y}^{2} = \frac{( {6 + 3cos \theta})^{2} + ( { - 3sin \theta})^{2} }{( {5 + 4cos \theta})^{2} } \\ = \frac{36 + 36cos \theta + 9 {cos}^{2} \theta + 9 {sin}^{2} \theta}{( {5 + 4cos \theta})^{2} } \\ = \frac{45 + 36cos \theta}{( {5 + 4cos \theta})^{2} } \\ = \frac{9}{5 + 4cos \theta} \\ = \frac{9 + 12cos \theta - 12cos \theta}{5 + 4cos \theta} \\ = \frac{24 + 12cos \theta - 15 - 12cos \theta}{5 + 4cos \theta} \\ = \frac{4(6 + 3cos \theta) - 3(5 + 4cos \theta)}{5 + 4cos \theta} \\ = \frac{4(6 + 3cos \theta)}{5 + 4cos \theta} - 3 \\ = 4x - 3 \\ \therefore \: \: \: \: \: \: \implies {x}^{2} + {y}^{2} = 4x - 3$

❀◕ ‿ ◕❀Hope it helps❀◕ ‿ ◕❀