Question

{i^18+(1/i)^24}^3 =0​

Answer 1

In complex numbers

i = √-1

i² = - 1, i³ = - i and i⁴ = 1

Again i⁵ = i and moves in this cycle.

For every four consecutive powers of i, the value repeats.

So,

i = i^(4n + 1) [n is any whole number]

i² = i^(4n+2) = -1

i³ = i^(4n+3) = -i

i⁴ = i^(4n)

Here

i^18 = i² = -1

i^24 = i⁴ = 1

So,

[i^18 + 1/(i^24)]³ = 0

⇒ [-1 + 1/(1)]³ = 0

⇒ [-1 + 1]³ = 0

⇒ 0³ =0

L.H.S = R.H.S

Hence proved.