if (a+ib) (c+id)(x+iy)=m+in
Answers
Answer:
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Answer:
It is given that x+iy = \sqrt{\frac{a+ib}{c+id}}x+iy=
c+id
a+ib
(Equation 1)
We have to prove that (x^2+y^2)^2 = \frac{a^2+b^2}{c^2+d^2}(x
2
+y
2
)
2
=
c
2
+d
2
a
2
+b
2
Consider the conjugate of x+iy,
So, x-iy = \sqrt{\frac{a-ib}{c-id}}x−iy=
c−id
a−ib
(Equation 2)
Multiplying equation 1 by 2, we get
(x+iy)(x-iy) =\sqrt{\frac{a+ib}{c+id}} \sqrt{\frac{a-ib}{c-id}}(x+iy)(x−iy)=
c+id
a+ib
c−id
a−ib
x^2+y^2 =\sqrt{\frac{(a+ib)(a-ib)}{(c+id)(c-id)}
x^2+y^2 =\sqrt{\frac{a^2-(ib)^2}{c^2-(id)^2}}x
2
+y
2
=
c
2
−(id)
2
a
2
−(ib)
2
x^2+y^2 =\sqrt{\frac{a^2+b^2}{c^2+d^2}}x
2
+y
2
=
c
2
+d
2
a
2
+b
2
Squaring on both the sides, we get
(x^2+y^2)^2 =\sqrt{(\frac{a^2+b^2}{c^2+d^2})}^2(x
2
+y
2
)
2
=
(
c
2
+d
2
a
2
+b
2
)
2
Therefore, (x^2+y^2)^2 =(\frac{a^2+b^2}{c^2+d^2})(x
2
+y
2
)
2
=(
c
2
+d
2
a
2
+b
2
)
Hence, proved.