randRange(1, 4) ['1', 'i', '-1', '-i'][EXP % 4]

`\large{i ^ {EXP} = ?}`

`SOLUTION`

- `1`
- `i`
- `-1`
- `-i`

Anything to the first power is the number itself.

The most important property of the imaginary unit `i` is that `\blue{i ^ 2} = \pink{-1}`.

`\qquad \begin{eqnarray} i ^ 3 &=& (\pink{i ^ 2}) \cdot i \\ &=& (\blue{-1}) \cdot i \\ &=& -i \end{eqnarray}`

`\qquad \begin{eqnarray} i ^ 4 &=& (\pink{i ^ 2}) ^ 2 \\ &=& (\blue{-1}) ^ 2 \\ &=& 1 \end{eqnarray}`

`\qquad i ^ EXP = SOLUTION`

4 + randRange(1, 30) ['1', 'i', '-1', '-i'][EXP % 4] floor(EXP / 4) EXP % 4 $._("remainder")

`\large{i ^ {EXP} = ?}`

`SOLUTION`

- `1`
- `i`
- `-1`
- `-i`

The most important property of the imaginary unit `i` is:
`\qquad \blue{i ^ 2} = \pink{-1}`

Therefore:
`\qquad i ^ 4 = (\blue{i ^ 2}) ^ 2 = (\pink{-1}) ^ 2 = 1`

So, we can simplify the expression by rewriting it in terms of `i^4`.

Because `EXP \div \blue{4} = \green{WHOLES} \text{ REMAINDER_TEXT } \red{REMAINDER}`,

`\qquad \begin{eqnarray} i ^ {EXP} &=& (i^\blue{4})^\green{WHOLES} \cdot i^\red{REMAINDER} \\ &=& (1)^\green{WHOLES} \cdot i^\red{REMAINDER} \\ &=& i^\red{REMAINDER} 1 \end{eqnarray}`

Anything to the first power is the number itself.
`\qquad i ^ \red{1} = i`

As stated above, `\blue{i ^ 2} = \pink{-1}`.

`\qquad \begin{eqnarray} i ^ \red{3} &=& (\blue{i ^ 2}) \cdot i \\ &=& (\pink{-1}) \cdot i \\ &=& -i \end{eqnarray}`

`i ^ {EXP} = i ^ {REMAINDER} = SOLUTION`