Power of i

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.

i²³

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

The remainder is 3.

Then i²³ = i³.

i³ = -i

So -i is the answer.

i⁸⁶

Divide 86 by 4.

The remainder is 2.

Then i⁸⁶ = i².

i² = -1

So -1 is the answer.

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

So -1 is the answer.