What is i^532 - 45?

i¹ = i
i² = -1
i³ = -i
i⁴ = 1
i⁵ = i

1 - 45
-44

Usually in mathematics, i is used as the root(-1). If you take the square root of this, you will get -1. The cube root will be root(-1) ^3 = root(-1) * root(-1)^2 = -1 * root(-1) = -i, If you do this for the fourth and fifth power, you will notice that every even power will give you -1, and every odd power will give you +-i.

In our case we have an even number, so i^(532) = -1.

i^(532) - 45 = -1 - 45 = -46

Does this refer to i = sqrt(-1)?

Powers of i go in cycles.

i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
i^5 = i
i^6 = -1, etc.

Figure out where 532 is in the cycle, by figuring out 532 mod 4 (remainder after division by 4). i^532 will be either i, -1, -i, or 1.