29Sir .. I tried the question like this..
x + iy = \frac{3}{(2+cos\theta ) + isin\theta } ...1
Taking bar on both sides
x - iy = \frac{3}{(2+cos\theta ) - isin\theta } ...2
Multiplying 1 and 2..
we will get
x^{2} +y^{2} = \frac{9}{(2+cos\theta )^{2} +sin^{2}\theta } = \frac{9}{5+ 4cos\theta }
and
x + iy = \frac{3}{(2+cos\theta ) + isin\theta }\frac{(2+cos\theta ) - isin\theta }{(2+cos\theta ) - isin\theta } = \frac{3(2+cos\theta ) -3 isin\theta }{5 + 4cos\theta }
so
x = \frac{3(2+cos\theta ) }{5 + 4cos\theta }
Now 4x - (x2 +y2) can be easily evaluated from the abv equations...
21i get a feeling tat i hav seen this somewer
anyways...here is teh soln.
\\\textup{let} \; z=x+iy=\frac{3}{cos\theta+isin\theta+2}\\ \Rightarrow cos\theta+isin\theta=\frac{3}{z}-2\\ \textup{Now take mod both sides}\\ 1=\left | \frac{3}{z} -2\right |\\ \Rightarrow \left | z\right |=2\left | z-\frac{3}{2}\right |\\ \Rightarrow \sqrt{x^2+y^2}=2\sqrt{\left ( x-\frac{3}{2} \right )^2+y^2}\\ \textup{squaring both sides}\\x^2+y^2=4(x^2+9/4-3x+y^2)\\ so\\ 4x-x^2-y^2=3
edited**actually user "∫" in post#2 did the same thing....my bad not to hav seen it.
341Good, you guys did much better than what's in that material.
I went about it in a more predictable manner perhaps than the solns given here:
x+iy = \frac{3}{\cos \theta + i \sin \theta+2}
\Rightarrow (x-2)+iy = \frac{3}{\cos \theta + i \sin \theta+2}-2 = - \frac{1+2\cos \theta + 2i \sin \theta}{\cos \theta + i \sin \theta + 2}
\Rightarrow (x-2)^2+y^2 = |(x-2)+iy|^2 = = \frac{|1+2\cos \theta + 2i \sin \theta|^2}{|\cos \theta + i \sin \theta + 2|^2} \\ \\ = \frac{5 + 4 \cos \theta}{5 + 4 \cos \theta} = 1
from which it follows that the given expression equals 3
