I think the key is cos(θ)+i sin(θ). Set that equal to i, -1, -i and 1 to figure out the angles.
Hi, Sorry if this is not in the right section - I did look but couldn't be sure.
I am trying to find all possible values of i^n
I know that the angle of i is pi/2 therefore there will be four possible values, being i, -1, -i and 1 as the number is rotated around, however I don't know how to really show this mathematically.
Any advice would be appreciated, thanks.
Since andy000 said "there will be four possible values, being i, -1, -i and 1" it is clear that he intended n to be an integer.
Frankly, I wouldn't use D'Moivre. I would just note that , , , . Since we are at "1" again and each step just multiplies by i again that will repeat. , , , and for any integer n. If you want a more "formal" way of showing that, note that so you really only need the first four values.
Your description lacks formality, but that is really all there is to it.
Let's clean this up a bit:
Suppose n = 4k + m, where m = 0,1,2 or 3. Every integer n falls into exactly one of these four categories.
Now $i^0 = 1$, by definition (we define $z^0 = 1$ for any non-zero complex number $z$, and $i$ is certainly not 0).
Also, $i^1 = i$. This is clear.
From the definition of $i$ (as a square root of -1), we have: $i^2 = -1$. So far, no heavy lifting at all.
$i^3 = (i^2)(i) = (-1)(i) = -i$.
$i^4 = (i^2)(i^2) = (-1)(-1) = 1$ <---this is what we will use from here on out.
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So, for ANY integer $k$, we have:
$i^{4k} = (i^4)^k = 1^k = 1$
$i^{4k+1} = (i^{4k})(i) = (1)(i) = i$ (see above)
$i^{4k+2} = (i^{4k})(i^2) = (1)(-1) = -1$
$i^{4k+3} = (i^{4k})(i^3) = (1)(-i) = -i$
....and that is all.
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Your intuition is correct, multiplying by $i$ corresponds to rotating (counter-clockwise) by $\dfrac{\pi}{2}$, which is a "quarter-turn". Four quarter-turns makes a whole turn, and we are back where we started, and repeat.
I did this for integers, rather than natural numbers, because:
$i^{-1} = \dfrac{1}{i} = \dfrac{1}{i}\cdot \dfrac{i}{i} = \dfrac{i}{i^2} = \dfrac{i}{-1} = -i$.
So $i^{-1} = i^3$ (note that -1 and 3 are "four apart"). This corresponds to our natural intuition that "a quarter turn backwards = three quarters of a turn forwards"
Notice the prominent role the number 4 plays, here. As we have seen, $i^4 = 1$, which means $i$ is a solution to $x^4 - 1 = 0$.
This is not surprising, because $x^4 - 1 = (x^2 + 1)(x^2 - 1)$ and $i$ is a root of $x^2 + 1$ (if $x^2 + 1 = 0$ then $x^2 = -1$, so $x$ is a square root of -1).
We have also seen $(-i)^4 = (-i)^2(-i)^2 = (-i)(-i(-i)(-i) = (-1)(i)(-1)(i)(-1)(i)(-1)(i) = (-1)^4i^4 = (1)(1) = 1$. So $-i$ is also a root of $x^4 - 1$.
This is ALSO not surprising as $x^2 + 1 = (x - i)(x + i) = (x - i)(x - (-i))$
Since -1 and 1 are the two roots of $x^2 - 1 = (x - 1)(x + 1)$, we conclude that $\{1,-1,i,-i\}$ are indeed ALL FOUR fourth roots of one (the four roots of $x^4 - 1$).
Note the various ways 4 figures in this:
$x^4 - 1$ <--a polynomial of degree 4
$\frac{\pi}{2}$ <--1/4th of a whole turn
$\{1,-1,i,-i\}$ <--a set of 4 things
This is no coincidence. There's a pattern, here.