Question

If x+iy=(a+ib)(c+id) where a,b,c,d,x and y are real numbers, prove that, X^2 + y^2= (ac-bd)^2 + (ac+bd)^2

Answer 1

Solution :

Method 1.

Given,

x + iy = (a + ib) (c + id)

⇒ x + iy = ac + iad + ibc + i²bd

⇒ x + iy = ac + iad + ibc - bd [ ∵ i² = - 1 ]

⇒ x + iy = (ac - bd) + i (ad + bc) ...(1)

Taking absolute value or modulus of the complex numbers from both the sides, we get

√(x² + y²) = √{(ac - bd)² + (ad + bc)²}

⇒ x² + y² = (ac - bd)² + (ad + bc)²

Hence, proved.

Method 2.

Comparing among the real and imaginary parts from both sides of (1), we get

x = ac - bd & y = ad + bc

∴ x² + y² = (ac - bd)² + (ad + bc)²

Hence, proved.