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If x+iy=3/(2+costheta +i sin theta), the...

If `x+iy=3/(2+costheta +i sin theta)`, then show that `x^2+y^2=4x-3`

Text Solution

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`3/(2+costheta+isintheta) = 3/(2+costheta+isintheta)**((2+costheta-isintheta))/((2+costheta-isintheta))`
`=(3(2+costheta-isintheta))/((2+costheta)^2-i^2sin^2theta)`
`=(6+3costheta-3isintheta)/(4+cos^2theta+4costheta+sin^2theta)`
`=((6+3costheta)-3isintheta)/(5+4costheta)`
`:. x+iy = ((6+3costheta)-3isintheta)/(5+4costheta)`

`=> x = (6+3costheta)/(5+4costheta), y = (-3sintheta)/(5+4costheta)`
Now. `L.H.S. = x^2+y^2 = ((6+3costheta)/(5+4costheta))^2+( (-3sintheta)/(5+4costheta))^2`
`=(36+9cos^2theta+36costheta+9sin^2theta)/(5+4costheta)^2`
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