Question

How do you simplify `i^101`?

Answer 1

Factor out multiples of `i^4`, yields dividing by `i^100` and obtaining `i`

Explanation:

Recall the rules of `i`. `i` is defined such that:

`i = i`
`i^2 = -1`
`i^3 = i*i^2 = -i`
`i^4 = (i^2)^2 = (-1)^2 = 1`

We also know that exponent rules function such that `(x^a)/(x^b) = x^(a-b)`. If we divide `i^101` by some factor of `i^101`, such as `i^x`, where `0<=x<101` and where `x` is divisible by 4, we will yield `i^(101-x)`

With this in mind, return to the initial problem. What is the largest power of `i` we can factor out of `i^101` (e.g. what is the highest `x` divisible by 4 and less than 101? This should be obtainable by subtracting 1 from 101 as many times as it takes to yield a number divisible by 4. As it turns out, we reach that point at `x=100`.

From here, the calculation is simple.

`i^101 = (i^101)/1 = (i^101)/(i^100) = i^(101-100) = i^1 = i`