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(b+ia)⁵ in Terms of (a+ib)⁵=α+iβ | Complex Numbers JEE

JEE Maths question with a full step-by-step solution.

Question

(a + ib)^5 = \alpha + i\beta

, then

(b + ia)^5

is equal to

\beta - i\alpha

\beta + i\alpha

correct

\alpha - \beta

-\alpha - i\beta

Solution

Step 1: Express

b + ia

as a multiple of the conjugate-related quantity

a - ib

. Multiply

a - ib

i

i(a - ib) = ia - i^2 b = ia + b = b + ia

b + ia = i(a - ib)

. Step 2: Raise to the fifth power:

(b + ia)^5 = \big[i(a - ib)\big]^5 = i^5 (a - ib)^5

Since

i^5 = i^4\cdot i = 1\cdot i = i

(b + ia)^5 = i\,(a - ib)^5

Step 3: Relate

(a - ib)^5

to the given quantity by conjugation. Because

a, b

are real,

a - ib = \overline{a + ib}

, and the conjugate of a power is the power of the conjugate:

(a - ib)^5 = \overline{(a + ib)^5} = \overline{\alpha + i\beta} = \alpha - i\beta

Step 4: Substitute back:

(b + ia)^5 = i(\alpha - i\beta) = i\alpha - i^2\beta = i\alpha + \beta = \beta + i\alpha

Correct answer: (2)

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