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

Question

#2/(3+costheta+isintheta)#

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Answer 1 · 2017-12-30T08:42:08

#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 · 2017-12-31T05:58:20

Please refer to a Proof given in the 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.!