Question

x+iy=1/a+ib prove that (x^2+y^2) (a^2+b^2)=1
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Answer 1

Answer:

Given : x+iy=1/a+ib

To Prove :  (x^2+y^2) (a^2+b^2)=1

Proof :

=> x+iy=1/a+ib

=> (x+iy)*(a+ib) = 1 ———————(i)

Conjugate of (x+iy) and (a+ib) is (x-iy) and (a-ib).

So,

=> (x-iy)*(a-ib) = 1  ———————(ii)

Multiplying (i) and (ii).

=> (x+iy)*(a+ib)*(x-iy)*(a-ib) = 1*1

=> (x+iy)*(x-iy) * (a+ib)*(a-ib) = 1

We know that (a+b)(a-b) = a^2 - b^2

So,

[x^2 - (iy)^2] * [a^2 - (ib)^2] = 1

[x^2 - i^2.y^2] * [a^2 - i^2.b^2] = 1

As i^2 = -1,

[x^2 + y^2] * [a^2 + b^2] = 1

HENCE PROVED!!!!

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Answer 2

Answer: Value of  (x^2+y^2) (a^2+b^2) is equal to 1

Step-by-step explanation:

$x+ \iota y = \frac{1}{a + \iota b}$

⇒ $(x+ \iota y)(a + \iota b)=1$

When two complex numbers are multiplied, the magnitude of the resulting complex number is equal to the product of magnitudes of the multiplied complex numbers.

Let z₁, z₂ and z₃ be 3 complex numbers such that z₁z₂ = z₃.

Then,

| z₁ | × | z₂ | = | z₃ |

Magnitude of complex number of form  p + iq is given by $\sqrt{p^{2} +q^{2} }$

$(x+ \iota y)(a + \iota b)=1$

Therefore,

⇒ $| x+ \iota y|* | a+ \iota b| = | 1 |$

⇒ $\sqrt{x^{2}+y^{2} }*\sqrt{a^{2}+b^{2} } = 1$

Squaring both sides,

$(x^{2} +y^{2} )(a^{2} +b^{2} ) = 1$

LHS = RHS

Hence, proved.

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