Question

a + ib = x +iy, prove that x2 + y2 etid a2 +62 c² + d²​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

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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