Quickly Calculate $\text i^{n}$ Values

[Drain Brain]

can you teach me how to quickly determine the values of the different powers of $\text i$ e.g $\text i^{587}$?

regards!

 

[Nono713]

You only need to remember that $i^0 = 1$, $i^1 = i$, $i^2 = -1$, $i^3 = -i$ (you really only need $i^2 = -1$, everything else is easily derivable) and that $i^4 = (i^2)^2 = (-1)^2 = 1$. That last relation means that if you are given a power of $i$, you can multiply or divide it by $i^4$, which does not change the result since $i^4$ is just one, but let's you add or subtract $4$ to/from the exponent. That means adding or subtracting 4 to the exponent makes no difference:
$$i^3 = i^7 = i^{11} = \cdots$$
$$i^3 = i^{-1} = i^{-5} = \cdots$$
Therefore the only thing that matters is the remainder of the exponent when divided by $4$:
$$i^k = \begin{cases} 1 ~ ~ \text{if} ~ k ~ \text{leaves a remainder of 0 when divided by} ~ 4 \\ i ~ ~ \text{if} ~ k ~ \text{leaves a remainder of 1 when divided by} ~ 4 \\ -1 ~ ~ \text{if} ~ k ~ \text{leaves a remainder of 2 when divided by} ~ 4 \\ -i ~ ~ \text{if} ~ k ~ \text{leaves a remainder of 3 when divided by} ~ 4 \end{cases}$$​

 

[Deveno]

The powers of $i$ form what professional mathematicians call a 4-cycle, which means exactly what you think it means. So:

$i^0 = 1$ (this is sort of "by default")
$i^1 = i$
$i^2 = -1$
$i^3 = (i^2)i = (-1)i = -i$
$i^4 = (i^2)(i^2) = (-1)(-1) = 1$, and we have come "full cycle" and it just repeats forever more after this.

Multiplying by $i$ can be thought of as "rotating counter-clockwise by a quarter-turn" (this should make sense, since we repeat every 4 1/4-turns), and thus multiplying by -1 can be thought of as an "about face" (a "180" as skaters like to call it), which perhaps (finally!) explains the curious rule that:

negative*negative = positive.

Therefore:

$i^{587} = i^{4\ast146 + 3} = (i^{4\ast146})(i^3) = (i^4)^{146}(i^3) = (1^{146})(i^3) = (1)(i^3) = i^3 = -i$