Powers of i

One more factor of i each time

Each power of i is the one before it multiplied by i. Start with $i^1=i$i¹ = i and $i^2=-1$i² = −1, the definition.

Then $i^3=i^2\times i=-1\times i=-i$i³ = i² × i = −1 × i = −i. And $i^4=i^3\times i=-i\times i=-i^2=-\left(-1\right)=1$i⁴ = i³ × i = −i × i = −i² = −(−1) = 1. The same comes from $i^4=i^2\times i^2=\left(-1\right)\times \left(-1\right)=1$i⁴ = i² × i² = (−1) × (−1) = 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 $i^4=1$i⁴ = 1 the powers start again: $i^5=i^4\times i=1\times i=i$i⁵ = i⁴ × i = 1 × i = i, then $i^6=-1$i⁶ = −1, $i^7=-i$i⁷ = −i and $i^8=1$i⁸ = 1. The values run i, −1, −i, 1 and then repeat in that order.

$i^0=1$i⁰ = 1, as for any nonzero number to the power 0, and it sits in the cycle where it should: $i^0=1$i⁰ = 1, $i^1=i$i¹ = i, $i^2=-1$i² = −1, $i^3=-i$i³ = −i, $i^4=1$i⁴ = 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 $i^4=1$i⁴ = 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 $i^{47}$i⁴⁷: 47 = 4 × 11 + 3, so $i^{47}={\left(i^4\right)}^{11}\times i^3=1^{11}\times \left(-i\right)=-i$i⁴⁷ = (i⁴)¹¹ × i³ = 1¹¹ × (−i) = −i. For $i^{41}$i⁴¹: 41 = 4 × 10 + 1, so $i^{41}=i$i⁴¹ = i. For $i^{2027}$i²⁰²⁷: 2027 = 4 × 506 + 3, so $i^{2027}=-i$i²⁰²⁷ = −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. $i^{10}$i¹⁰ is two full cycles and then $i^2$i², so it is −1.

Four in a row add to 0

$i+i^2+i^3+i^4=i-1-i+1=0$i + i² + i³ + i⁴ = 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 $i+i^2+i^3+\cdots +i^{10}$i + i² + i³ + … + i¹⁰, the first eight terms make two blocks that each add to 0, and the sum is $i^9+i^{10}=i+\left(-1\right)=-1+i$i⁹ + i¹⁰ = i + (−1) = −1 + i.

Negative powers

$i^{-1}=\frac{1}{i}$i⁻¹ = 1/i, the number that multiplies i to give 1. Since $i\times \left(-i\right)=-i^2=1$i × (−i) = −i² = 1, $i^{-1}=-i$i⁻¹ = −i.

Going backward round the cycle divides by i each time: $i^{-1}=-i$i⁻¹ = −i, $i^{-2}=-1$i⁻² = −1, $i^{-3}=i$i⁻³ = i and $i^{-4}=1$i⁻⁴ = 1. The remainder rule still works, with the remainder taken between 0 and 3: −1 = 4 × (−1) + 3, so $i^{-1}=i^3=-i$i⁻¹ = i³ = −i, and −6 = 4 × (−2) + 2, so $i^{-6}=i^2=-1$i⁻⁶ = i² = −1.

The same ring walked the other way, dividing by i: 1, then −i, then −1, then i, and back to 1. These are $i^0$i⁰, $i^{-1}$i⁻¹, $i^{-2}$i⁻², $i^{-3}$i⁻³ and $i^{-4}$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 $i^{41}=i$i⁴¹ = i.

Stopping one step short or going one too far. $i^3$i³ is −i, not −1 (that is $i^2)$i²) and not i (that is $i^5)$i⁵).

Taking $i^4$i⁴ as −1. Two factors of i give −1, so four give (−1) × (−1) = 1.

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