Dear Student, Solution: Given: (a+ib)5=α+iβor i5(b-ai)5=α+iβ {Taking i5 common from RHS}⇒i(b-ai)5=α+iβ {Since i4=1}⇒(b-ai)5=β-iα{Multipyling both sides with -i, -i2=1}Now, taking conjugate both sides we get(b+ia)5=β+iα {since z5¯=z¯5} Answer is option A Regards

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Dear Student,
Solution:

Given : {(a + ib)}^{5} = α + iβ or i^{5} {(b - ai)}^{5} = α + iβ {Taking i^{5} common from RHS} \Rightarrow i {(b - ai)}^{5} = α + iβ {Since i^{4} = 1} \Rightarrow {(b - ai)}^{5} = β - iα {Multipyling both sides with - i, - i^{2} = 1} Now, taking conjugate both sides we get {(b + ia)}^{5} = β + iα {since \bar{z^{5}} = {\bar{z}}^{5}}

Answer is option A.
Regards

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