Question

x+iy=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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