Question

Give Explanation for your Answer (Q.22 )

Answer 1

hey mate
here's the solution

Answer 2

(22)

Given Equation is (x + iy)^(1/3) = (a + ib)

On cubing both sides, we get

⇒ (x + iy) = (a + ib)^3

We know that (a + b)^3 = a^3 + b^3 + 3ab(a + b).

               = a^3 + (ib)^3 + 3(a)(ib)(a + ib)

               = a^3 + i^3b^3 + 3aib(a + ib)

               = a^3 + (i^2)(i)b^3 + 3a^2ib + 3ai^2b^2

We know that i^2 = -1

               = a^3 - ib^3 + 3a^2ib - 3ab^2

               = a^3 - 3ab^2 + 3a^2ib - ib^3

               = a^3 - 3ab^2 + i(3a^2b - b^3)

(x + iy)     = a^3 - 3ab^2 + i(3a^2b - b^3)

Equating real and imaginary parts on both sides, we get

⇒ x = a^3 - 3ab^2

⇒ y = 3a^2b - b^3.

Now,

⇒ (x/a) - (y/b)

⇒ (a^3 - 3ab^2/(a)) - (3a^2b - b^3/(b))

⇒ a^2 - 3b^2 - 3a^2 + b^2

⇒ -2a^2 - 2b^2

⇒ -2(a^2 + b^2)

Hope it helps!