Question

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

`2/(3+costheta+isintheta)`

Answer 1

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

Answer 2

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.!