if x+iy=2÷(3+cos theta + i sin theta) ; prove that 2x^2+2y^2= 3x - 1
Answers
Answer:
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Step-by-step explanation:
We have x+iy=
(2+cosθ)+isinθ
3
Rationalizing the denominator, we get
x+iy=
((2+cosθ)+isinθ)
3
((2+cosθ)−isinθ)
((2+cosθ)−isinθ)
∴x+iy=
((2+cosθ)
2
−(isinθ)
2
)
3((2+cosθ)−isinθ)
∴x+iy=
4+cos
2
θ+4cosθ+sin
2
θ
6+3cosθ−3isinθ
∴x+iy=
5+4cosθ
6+3cosθ−3isinθ
∴x=
5+4cosθ
6+3cosθ
and y=−
5+4cosθ
3sinθ
∴x
2
+y
2
=
(5+4cosθ)
2
(6+3cosθ)
2
+(3sinθ)
2
∴x
2
+y
2
=
(5+4cosθ)
2
36+36cosθ+9cos
2
θ+9sin
2
θ
∴x
2
+y
2
=
(5+4cosθ)
2
45+36cosθ
∴x
2
+y
2
=
(5+4cosθ)
2
9(5+4cosθ)
∴x
2
+y
2
=
(5+4cosθ)
9
∴x
2
+y
2
=
(5+4cosθ)
9+12cosθ−12cosθ
∴x
2
+y
2
=
(5+4cosθ)
24+12cosθ−15−12cosθ
∴x
2
+y
2
=
(5+4cosθ)
4(6+3cosθ)−3(5+4cosθ)
∴x
2
+y
2
=4
(5+4cosθ)
(6+3cosθ)
−3
∴x
2
+y
2
=4x−3
Hence, proved.