Question

What is the answer?

Imaginary numbers:

Answer 1

`i^0 = 1`
`i^1 = i`
`i^2 = -1`
`i^3 = -i`
`i^4 = 1`

Explanation:

Generally if you want to get the solution to `i^x` where `x` is any non-negative integer, you need to remember the following:

`i^0 = 1`
`i^1 = i`
`i^2 = -1`
`i^3 = -i`
`i^4 = 1`

For all non-negative integers, if it is an exponent of `i` then you simply have to divide `x` by 4, 3, or 2.

If `x` is divisible by 4 then `i^x = 1`
If `x` is divisible by 3 then `i^x = -i`
If `x` is divisible by 2 then `i^x = -1`
If `x` is not divisible by any of the above numbers then `i^x = i`
If `x` is 0 then `i^x = 1` since anything raised to 0 is equal to 1
Note that for this you must choose the highest number among 4, 3, and 2 that `x` is divisible with.

Similarly,
`i^2 = i * i = -1`, so it also goes that:
`i^3 = i * i * i = i^2 * i = -1 * i = -i`
`i^4 = i * i * i * i = i^2 * i^2 = -1 * -1 = 1`

Answer 2

`i^0=1`
`i^1=i=sqrt(-1)`
`i^2=-1`
`i^3=-i=-sqrt(-1)`
`i^4=1`

Explanation:

`i=sqrt(-1)`

`i^0=(x^0=1 | x in C)` ... any number (x) power 0 is 1
`i^1=i` ... any number (x) power 1 is the number (x)

`i^2=i*i=(sqrt(-1))(sqrt(-1))=-1`

`i^3=i*i^2=sqrt(-1)*1=sqrt(-1)=i`

`i^4=i^(2+2)=i^2*i^2=(-1)*(-1)=1`

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note: for any `i^a=b`, one can write
`(i^(2n))(i^(a-2n))=b`
when
`2n` is Even
AND
`2n-a=1 or 2n-a=3`

for exmple:
`i^35=?`

We will take `(2n=32,2n-a=3)`

`i^35=i^32*i^3=((i^4)^8)*i^3=(1)^8*i^3=1*i^3=i^3=i`

end note