Powers of i

One more factor of i each time

Each power of i is the one before it multiplied by i. Start with $i^{1} = i$ and $i^{2} = -1$, the definition.

Then $i^{3} = i^{2} \times i = -1 \times i = -i$. And $i^{4} = i^{3} \times i = -i \times i = -i^{2} = -\left(-1\right) = 1$. The same comes from $i^{4} = i^{2} \times i^{2} = \left(-1\right) \times \left(-1\right) = 1$.

So four factors of i make 1. Each factor of i is a quarter turn, as in The Imaginary Unit, and four quarter turns make a full turn back to the start.

The cycle of four

After $i^{4} = 1$ the powers start again: $i^{5} = i^{4} \times i = 1 \times i = i$, then $i^{6} = -1$, $i^{7} = -i$ and $i^{8} = 1$. The values run i, −1, −i, 1 and then repeat in that order.

$i^{0} = 1$, as for any nonzero number to the power 0, and it sits in the cycle where it should: $i^{0} = 1$, $i^{1} = i$, $i^{2} = -1$, $i^{3} = -i$, $i^{4} = 1$.

Each arrow multiplies by i: 1, then i, then −1, then −i, and back to 1. The power of i counts how many arrows have been followed from 1.

Only the remainder matters

Since $i^{4} = 1$, every block of four factors of i is 1 and can be dropped. Divide the power by 4: the quotient counts the full cycles, which contribute 1, and the remainder decides the value.

Remainder 0 gives 1, remainder 1 gives i, remainder 2 gives −1, and remainder 3 gives −i.

For $i^{47}$: 47 = 4 × 11 + 3, so $i^{47} = \left(i^{4}\right)^{11} \times i^{3} = 1^{11} \times \left(-i\right) = -i$. For $i^{41}$: 41 = 4 × 10 + 1, so $i^{41} = i$. For $i^{2027}$: 2027 = 4 × 506 + 3, so $i^{2027} = -i$.

The powers 0 to 12 along a line, with the value of i to that power marked above. Every multiple of 4 gives 1. $i^{10}$ is two full cycles and then $i^{2}$, so it is −1.

Four in a row add to 0

$i + i^{2} + i^{3} + i^{4} = i - 1 - i + 1 = 0$. The same holds for any four powers in a row, since they are the same four values in a different order.

So in $i + i^{2} + i^{3} + \dots + i^{10}$, the first eight terms make two blocks that each add to 0, and the sum is $i^{9} + i^{10} = i + \left(-1\right) = -1 + i$.

Negative powers

$i^{-1} = \frac{1}{i}$, the number that multiplies i to give 1. Since $i \times \left(-i\right) = -i^{2} = 1$, $i^{-1} = -i$.

Going backward round the cycle divides by i each time: $i^{-1} = -i$, $i^{-2} = -1$, $i^{-3} = i$ and $i^{-4} = 1$. The remainder rule still works, with the remainder taken between 0 and 3: −1 = 4 × (−1) + 3, so $i^{-1} = i^{3} = -i$, and −6 = 4 × (−2) + 2, so $i^{-6} = i^{2} = -1$.

The same ring walked the other way, dividing by i: 1, then −i, then −1, then i, and back to 1. These are $i^{0}$, $i^{-1}$, $i^{-2}$, $i^{-3}$ and $i^{-4}$.

The usual mistakes

Using the quotient instead of the remainder. 41 ÷ 4 = 10.25, and neither 10 nor 0.25 is what matters: the remainder is 1, so $i^{41} = i$.

Stopping one step short or going one too far. $i^{3}$ is −i, not −1 (that is $i^{2})$ and not i (that is $i^{5})$.

Taking $i^{4}$ as −1. Two factors of i give −1, so four give (−1) × (−1) = 1.

Taking $i^{-1}$ as i or −1. i × (−i) = 1, so $i^{-1} = -i$.