Powers of i go round in fours.

The point iⁿ hops a quarter turn per step round the unit circle, landing on 1, i, −1, −i; a counter shows n mod 4 and a running sum of four consecutive powers.

Set n = 103 and read i¹⁰³ = −i; then set n = −22 and confirm it lands on −1.

Taking i³ = i or 1.

• The powers of i repeat in a cycle of four: i¹ = i, i² = −1, i³ = i² × i = −i, i⁴ = (i²)² = 1, and then i⁵ = i again.
• For every integer k: i⁴ᵏ = 1, i⁴ᵏ⁺¹ = i, i⁴ᵏ⁺² = −1 and i⁴ᵏ⁺³ = −i.
• To find iⁿ, divide n by 4 and keep only the remainder. For n = 103 = 4 × 25 + 3, i¹⁰³ = i³ = −i; for n = 58 = 4 × 14 + 2, i⁵⁸ = −1.
• Negative powers follow the same cycle: i⁻¹ = 1/i = −i, i⁻² = −1, i⁻³ = i and i⁻⁴ = 1. For i⁻²², write −22 = 4 × (−6) + 2, so i⁻²² = i² = −1; directly, 1/i²² = 1/(−1) = −1.
• Any four consecutive powers add to zero: iⁿ + iⁿ⁺¹ + iⁿ⁺² + iⁿ⁺³ = iⁿ(1 + i − 1 − i) = 0.
• Seen on the Argand plane, each extra factor of i is a quarter turn anticlockwise, so four factors make a full turn and bring the point back to where it started.

Teaching this idea? Put the simulation on your own page or LMS with this code.