If x+iy=a+ib/c+id prove that x^2+y^2=1
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Given that,
x− iy=
c−id
a−ib
⟹ (x− iy)
2
=
c−id
a−ib
×
c+id
c+id
=
2
c 2
(ac+bd)−i(bc−ad)
⟹ (x
2
−
y 2
i(2xy)=
(
c 2 + d 2
a c + b d
)
− i
(
c 2 + d 2
b c − a d
)
Equating real and imaginary parts on both sides, we get
2xy=
x 2
y 2
c 2 + d 2
a c + b d
c 2 + d 2
b c − a d
Now,
(x+
iy)
2
=
(x
2
−
y 2
i(2xy)=
(
c 2 + d 2
a c + b d
)
+
i
(
c 2 + d 2
b c − a d
)
⟹ (x+ iy)
2
=
2
c 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
+
2
y 2
[(x− iy)(x+ iy)]
2
= (x− iy)
2
(x+
iy)
2
=
(
c − i d
a − i b
)
(
c + i d
a + i b
)
=
c 2 + d 2
a 2 + b 2
=
RHS
Hence proved.
Step-by-step explanation:
hope it's helpful for all
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