# Power of i

How to solve the power of i (i, i^{2}, i^{3}, i^{4}, ... ): formula, 3 examples, and their solutions.

## Formula

### Formula

i = i

i^{2}

= (√-1)^{2}

= -1

Imaginary Number i

i^{3}

= i^{2}⋅i

= -1⋅i

= -i

i^{4}

= (i^{2})^{2}

= (-1)^{2}

= +1

i = i

i^{2} = -1

i^{3} = -i

i^{4} = 1

i^{5}

= i^{4}⋅i

= 1⋅i

= i

i^{6}

= i^{4}⋅i^{2}

= 1⋅(-1)

= -1

i^{7}

= i^{4}⋅i^{3}

= 1⋅(-i)

= -i

i^{8}

= i^{4}⋅i^{4}

= 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

### Example

### Solution

i^{23}

Divide 23 by 4.

(4 is the cycle of the powers of i.)

The remainder is 3.

Then i^{23} = i^{3}.

i^{3} = -i

So -i is the answer.

## Example 2

### Example

### Solution

i^{86}

Divide 86 by 4.

The remainder is 2.

Then i^{86} = i^{2}.

i^{2} = -1

So -1 is the answer.

## Example 3

### Example

### Solution

The exponents goes from 1 to 99.

So there are 99 terms.

Then divide 99 by 4.

Group [i + i^{2} + i^{3} + i^{4}].

Group [i^{5} + i^{6} + i^{7} + i^{8}].

The quotient is 24.

So there are 24 of these groups.

The remainder is 3.

So write the last 3 terms:

+i^{97} + i^{98} + i^{99}.

i = i

i^{2} = -1

i^{3} = -i

i^{4} = 1

i^{5} = i

i^{6} = -1

i^{7} = -i

i^{8} = 1

Then, by the same way,

i^{97} = i

i^{98} = -1

i^{99} = -i.

So

[i + i^{2} + i^{3} + i^{4}] + [i^{5} + i^{6} + i^{7} + i^{8}] + ... + i^{97} + i^{98} + i^{99}

= [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

So -1 is the answer.