if x+iy √a+ib/c+ib , prove that (x²+y²) ² = a²+b²/c²+d²
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Step-by-step explanation:
Answer
Given that,
x−iy=
c−id
a−ib
⟹(x−iy)
2
=
c−id
a−ib
×
c+id
c+id
=
c
2
+d
2
(ac+bd)−i(bc−ad)
⟹(x
2
−y
2
)−i(2xy)=(
c
2
+d
2
ac+bd
)−i(
c
2
+d
2
bc−ad
)
Equating real and imaginary parts on both sides, we get
x
2
−y
2
=
c
2
+d
2
ac+bd
and 2xy=
c
2
+d
2
bc−ad
Now, (x+iy)
2
=(x
2
−y
2
)+i(2xy)=(
c
2
+d
2
ac+bd
)+i(
c
2
+d
2
bc−ad
)
⟹(x+iy)
2
=
c
2
+d
2
(ac+bd)+i(bc−ad)
=
(c+id)(c−id)
(a+ib)(c−id)
=
c+id
a+ib
⟹x+iy=
c+id
a+ib
LHS=(x
2
+y
2
)
2
=[(x−iy)(x+iy)]
2
=(x−iy)
2
(x+iy)
2
=(
c−id
a−ib
)(
c+id
a+ib
)
=
c
2
+d
2
a
2
+b
2
=RHS
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