The imaginary unit i: square roots of negative numbers, the four powers of i and why x² = −1 has no real solution
Square any real number and the answer is never negative: 3² = 9 and (−3)² = 9. So the equation x² = −1 has no real solution. You can see it on a graph: the parabola y = x² + 1 never touches the x-axis. Mathematicians solved this by inventing a new number i with i² = −1. It is called imaginary only as a name; it is as useful as any other number. Now square roots of negative numbers make sense: √(−9) = √9 × i = 3i, because (3i)² = 9 × i² = −9. In general √(−n) = √n × i for n > 0. The powers of i follow a cycle of four: i¹ = i, i² = −1, i³ = i² × i = −i, i⁴ = 1, and then i⁵ = i again. To find iⁿ, divide n by 4 and look at the remainder; this also works for negative n if you write n as a multiple of 4 plus a remainder from 0 to 3 (−45 = −48 + 3). Beware: √a × √b = √(ab) fails when both numbers are negative: √(−4) × √(−9) = 2i × 3i = 6i² = −6, not √36 = 6.
| Remainder of the exponent ÷ 4 | Value of iⁿ |
|---|---|
| 0 | 1 |
| 1 | i |
| 2 | −1 |
| 3 | −i |
Fill in the table. Copy the row of iⁿ for n = 1 to 8 on paper, then continue it up to n = 12 without a calculator. Which values come back after every four steps?