# Power of i

How to solve the power of i (i, i2, i3, i4, ... ): formula, 3 examples, and their solutions.

## Formula

### Formula

i = i

i2
= (√-1)2
= -1

Imaginary Number i

i3
= i2⋅i
= -1⋅i
= -i

i4
= (i2)2
= (-1)2
= +1

i = i
i2 = -1
i3 = -i
i4 = 1

i5
= i4⋅i
= 1⋅i
= i

i6
= i4⋅i2
= 1⋅(-1)
= -1

i7
= i4⋅i3
= 1⋅(-i)
= -i

i8
= i4⋅i4
= 1⋅1
= 1

As you can see,
i, -1, -i, +1
is repeating.

This is the way
to solve the powers of i.

## Example 1

### Solution

i23

Divide 23 by 4.
(4 is the cycle of the powers of i.)

The remainder is 3.

Then i23 = i3.

i3 = -i

## Example 2

### Solution

i86

Divide 86 by 4.

The remainder is 2.

Then i86 = i2.

i2 = -1

## Example 3

### Solution

The exponents goes from 1 to 99.
So there are 99 terms.

Then divide 99 by 4.

Group [i + i2 + i3 + i4].
Group [i5 + i6 + i7 + i8].

The quotient is 24.
So there are 24 of these groups.

The remainder is 3.
So write the last 3 terms:
+i97 + i98 + i99.

i = i
i2 = -1
i3 = -i
i4 = 1

i5 = i
i6 = -1
i7 = -i
i8 = 1

Then, by the same way,
i97 = i
i98 = -1
i99 = -i.

So
[i + i2 + i3 + i4] + [i5 + i6 + i7 + i8] + ... + i97 + i98 + i99
= [i - 1 - i + 1] + [i - 1 - i + 1] + ... + i - 1 - i.

i - 1 - i + 1 = 0
So each group is 0.

i - 1 - i = -1

0 + 0 + ... - 1 = -1