Question

If $x+iy=\frac{3}{2+\cos \theta +i\sin \theta }$ then, show that $x^2+y^2=4x-3$.

Answer 1

We have $x+iy=\frac{3}{(2+\cos \theta )+i\sin \theta }$

Rationalizing the denominator, we get

$x+iy=\frac{3}{\left(\right(2+\cos \theta )+i\sin \theta )}\frac{\left(\right(2+\cos \theta )-i\sin \theta )}{\left(\right(2+\cos \theta )-i\sin \theta )}$

$\therefore x+iy=\frac{3\left(\right(2+\cos \theta )-i\sin \theta )}{\left(\right(2+\cos \theta {)}^2-\left(i\sin \theta {)}^2\right)}$

$\therefore x+iy=\frac{6+3\cos \theta -3i\sin \theta }{4+{\cos }^2\theta +4\cos \theta +{\sin }^2\theta }$

$\therefore x+iy=\frac{6+3\cos \theta -3i\sin \theta }{5+4\cos \theta }$

$\therefore x=\frac{6+3\cos \theta }{5+4\cos \theta }$ and $y=-\frac{3\sin \theta }{5+4\cos \theta }$

$\therefore x^2+y^2=\frac{(6+3\cos \theta {)}^2+(3\sin \theta {)}^2}{(5+4\cos \theta {)}^2}$

$\therefore x^2+y^2=\frac{36+36\cos \theta +9{\cos }^2\theta +9{\sin }^2\theta }{(5+4\cos \theta {)}^2}$

$\therefore x^2+y^2=\frac{45+36\cos \theta }{(5+4\cos \theta {)}^2}$

$\therefore x^2+y^2=\frac{9(5+4\cos \theta )}{(5+4\cos \theta {)}^2}$

$\therefore x^2+y^2=\frac{9}{(5+4\cos \theta )}$

$\therefore x^2+y^2=\frac{9+12\cos \theta -12\cos \theta }{(5+4\cos \theta )}$

$\therefore x^2+y^2=\frac{24+12\cos \theta -15-12\cos \theta }{(5+4\cos \theta )}$

$\therefore x^2+y^2=\frac{4(6+3\cos \theta )-3(5+4\cos \theta )}{(5+4\cos \theta )}$

$\therefore x^2+y^2=4\frac{(6+3\cos \theta )}{(5+4\cos \theta )}-3$

$\therefore x^2+y^2=4x-3$

Hence, proved.