Evaluate:(i) \((\sqrt{-1})^{192}\)(ii) \((\sqrt{-1})^{93}\)(iii) \(

1.	Evaluate:(i) \((\sqrt{-1})^{192}\)(ii) \((\sqrt{-1})^{93}\)(iii) \((\sqrt{-1})^{30}\)
Answer» Since i = \((\sqrt{-1})\) so (i) L.H.S. = \((\sqrt{-1})^{192}\) ⇒ i¹⁹² ⇒ i^{4 x 48} = 1 Since it is of the form i⁴ⁿ = 1 so the solution would be 1 (ii) L.H.S.= \((\sqrt{-1})^{93}\) ⇒ i^{4 x 23 + 1} ⇒ i^{4n + 1} ⇒ i¹= i Since it is of the form of i^{4n + 1} = i so the solution would be simply i. (iii) L.H.S = \((\sqrt{-1})^{30}\) ⇒ i^{4 x 7 + 2} ⇒ i^{4n + 2} ⇒ i² = -1 Since it is of the form i^{4n + 2} so the solution would be -1

Evaluate:(i) \((\sqrt{-1})^{192}\)(ii) \((\sqrt{-1})^{93}\)(iii) \((\sqrt{-1})^{30}\)

Answer»

Since i = \((\sqrt{-1})\) so

(i) L.H.S. = \((\sqrt{-1})^{192}\)

⇒ i¹⁹²

⇒ i^{4 x 48} = 1

Since it is of the form i⁴ⁿ = 1 so the solution would be 1

(ii) L.H.S.= \((\sqrt{-1})^{93}\)

⇒ i^{4 x 23 + 1}

⇒ i^{4n + 1}

⇒ i¹= i

Since it is of the form of i^{4n + 1} = i so the solution would be simply i.

(iii) L.H.S = \((\sqrt{-1})^{30}\)

⇒ i^{4 x 7 + 2}

⇒ i^{4n + 2}

⇒ i² = -1

Since it is of the form i^{4n + 2} so the solution would be -1