If x+iy=2/(3+costheta+isintheta),show that 2x^2+2y^2=3x-1?

2/(3+costheta+isintheta)

2 Answers
Dec 30, 2017

2(x^2+y^2) is purely a real no., and, (3x-1) is a complex no. So, how these two can be equal?

Dec 31, 2017

Please refer to a Proof given in the Explanation.

Explanation:

Given that, x+iy=2/(3+costheta+isintheta).

:. (x+iy)(3+costheta+isintheta)=2, i.e.,

(3+costheta)x+ixsintheta+(3+costheta)iy+i^2ysintheta=2, or,,

(3+costheta)x-ysintheta+i{xsintheta+(3+costheta)y}=2+0i.

Comparing the Real and Imaginary Parts, we have,

(3+costheta)x-ysintheta=2, and, xsintheta+(3+costheta)y=0.

Solving these eqns. for (3+costheta) and sintheta, we get,

3+costheta=(2x)/(x^2+y^2) and sintheta=-(2y)/(x^2+y^2), or,

costheta=(2x)/(x^2+y^2)-3 and sintheta=-(2y)/(x^2+y^2).

But, cos^2theta+sin^2theta=1,

rArr {(2x)/(x^2+y^2)-3}^2+{-(2y)/(x^2+y^2)^2}=1.

rArr (4x^2)/(x^2+y^2)^2-(12x)/(x^2+y^2)+9+(4y^2)/(x^2+y^2)^2=1.

rArr (4x^2+4y^2)/(x^2+y^2)^2=(12x)/(x^2+y^2)-8, i.e.,

(4(x^2+y^2))/(x^2+y^2)^2=4{(3x)/(x^2+y^2)-2}, or,

1/(x^2+y^2)=(3x)/(x^2+y^2)-2.

rArr 2=(3x-1)/(x^2+y^2)," giving,"

2(x^2+y^2)=3x-1, as desired!

Q.E.D.

Enjoy Maths.!