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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
If (a+ib)5=α+iβ(a + ib)^5 = \alpha + i\beta, then (b+ia)5(b + ia)^5 is equal to
Aβ−iα\beta - i\alpha
Bβ+iα\beta + i\alphacorrect
Cα−β\alpha - \beta
D−α−iβ-\alpha - i\beta
Solution
Step 1: Express b+iab + ia as a multiple of the conjugate-related quantity a−iba - ib. Multiply a−iba - ib by ii:
i(a−ib)=ia−i2b=ia+b=b+iai(a - ib) = ia - i^2 b = ia + b = b + ia
 So b+ia=i(a−ib)b + ia = i(a - ib). Step 2: Raise to the fifth power:
(b+ia)5=[i(a−ib)]5=i5(a−ib)5(b + ia)^5 = \big[i(a - ib)\big]^5 = i^5 (a - ib)^5
 Since i5=i4⋅i=1⋅i=ii^5 = i^4\cdot i = 1\cdot i = i:
(b+ia)5=i (a−ib)5(b + ia)^5 = i\,(a - ib)^5
 Step 3: Relate (a−ib)5(a - ib)^5 to the given quantity by conjugation. Because a,ba, b are real, a−ib=a+ib‾a - ib = \overline{a + ib}, and the conjugate of a power is the power of the conjugate:
(a−ib)5=(a+ib)5‾=α+iβ‾=α−iβ(a - ib)^5 = \overline{(a + ib)^5} = \overline{\alpha + i\beta} = \alpha - i\beta
 Step 4: Substitute back:
(b+ia)5=i(α−iβ)=iα−i2β=iα+β=β+iα(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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