Question

if x+iy=3/(2+cos theta + i sin theta ) ; prove that x^2+y^2=4x-3

Answer 1

$\textbf{Given:}$

$x+iy=\frac{3}{2+cos\theta+i\;sin\theta}$

$\implies\;(x+iy)(2+cos\theta+i\;sin\theta)=3$

$\implies\;x(2+cos\theta)+i\;xsin\theta+i\;y(2+cos\theta)-ysin\theta=3$

$\implies\;(2x+xcos\theta-ysin\theta)+i(xsin\theta+2y+ycos\theta)=3+i\;0$

$\text{Separating corresponding real and imaginary parts on bothsides, we get}$

$2x+xcos\theta-ysin\theta=3$

$\text{and}$

$xsin\theta+2y+ycos\theta=0$

$\implies$

$xcos\theta-ysin\theta=3-2x$...(1)

$\text{and}$

$xsin\theta+ycos\theta=-2y$...(2)

$\text{Squaring and adding (1) and (2)}$

$(xcos\theta-ysin\theta)^2+(xsin\theta+ycos\theta)^2=(3-2x)^2+4y^2$

$x^2cos^2\theta+y^2sin^2\theta-2xy\;cos\theta\;sin\theta+x^2sin^2\theta+y^2cos^2\theta+2xy\;cos\theta\;sin\theta=9+4x^2-12x+4y^2$

$x^2(cos^2\theta+sin^2\theta)+y^2(cos^2\theta+sin^2\theta)=9+4x^2-12x+4y^2$

$\text{using,}\bf\;cos^2A+sin^2A=1$

$x^2+y^2=9+4x^2-12x+4y^2$

$\text{Rearranging terms, we get}$

$-3x^2-3y^2=9-12x$

$3x^2+3y^2=12x-9$

$\implies\boxed{\bf\;x^2+y^2=4x-3}$

Answer 2

Answer:

your answer attached in the photo