Question

Evaluate:

(i) i^–50

(ii) i^–9

(ii) i^–131.

Answer 1

(i) L.H.S. = i^–50

⇒ i^{-4 x 13 + 2}

⇒ i^{4n + 2}

⇒ -1

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

(ii) L.H.S. = i^–9

⇒ i^{-4 x 3 + 3}

⇒ i^{4n + 3}

⇒ i³ = -i

Since it is of the form of i^{4n + 3} so the solution would be simply -i.

(ii) L.H.S. = i^–131.

⇒ i^{-4 x 33 + 1}

⇒ i^{4n + 1}

⇒ i¹ = i

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