How is it possible to find *all* of the roots of a complex number? I was asked to find every fourth root of the complex number $i$.
Setting $z=i$, we get
$z=e^{i\frac{\pi}{2}}=e^{i(\frac{\pi}{2}+2\pi m)}$, where $m$ is any integer. 
$z^{\frac{1}{4}}=[=e^{i(\frac{\pi}{2}+2\pi m)}]^{\frac{1}{4}}$.
The fourth roots are then found by plugging integer values of $m$ into the above. Since this is the case, how is it that I can find all of the roots?
Surely you can plug in an infinite number of integers and get an infinite number of roots?
For example, using:
$$m=0, i^{\frac{1}{4}}=\frac{\pi}{8}$$
$$m=1, i^{\frac{1}{4}}=\frac{5\pi}{8}$$
$$m=2, i^{\frac{1}{4}}=\frac{9\pi}{8}$$
and so on, each value of $m$ yielding a different result.
 A: There are only four roots; putting in larger integers will give you repetitions of the four roots. This is because
$$e^{2\pi i}=1$$
A: A nonzero complex number has exactly $m$ $m$'th roots.
A: Note that by using periodicity of trigonometric functions or the exponent rule: $$e^{i(2\pi+\theta)}=e^{i2\pi}e^{i\theta}=e^{i\theta}\tag{1}$$
As your condition is:
$$e^{i4\theta}=i=e^{2n\pi+i\pi/2}\quad n\in\mathbb N$$
So:
$$4\theta=2n\pi+\pi/2\implies \theta=\frac{n\pi}2+\frac{\pi}{8}$$
Now you should have $\theta<2\pi$ otherwise we will get the same root using $(1)$. So:
$$\theta=\frac{(4n+1)\pi}8,n\in\{0,1,2,3\}$$
Since at $n=4$, $\theta=17\pi/8>2\pi$.
A: You're confusing polar representations of complex numbers with the numbers themselves. Just because $e^{i\theta} = e^{i(\theta + 2\pi k)}$ for all integers $k$, doesn't mean there are infinitely many values of $e^{i\theta}$, just infinitely many representations. E.g., $1 = 0 + 1 = -1 + 2 = -2 + 3 = \ldots$ are all just different ways of writing $1$.
More seriously, when you write $i^{1/4} = \frac{\pi}{8}$, you're confusing a complex number - one of several fourth roots of $i$ - with its argument, $\pi/8$. You should be writing $i^{1/4} = e^{i\pi/8}$. But even that's just better and not a good idea, as $i^{1/4}$ doesn't mean just one number, but a set of $4$ numbers.
Given a complex number $z = e^{i\theta}$ of size $1$, there are exactly $n$ $n$-th roots, the set $\{e^{i(\theta+2\pi k)/n} : k = 0, 1, \ldots, n-1\}$. Of course, you can add multiples of $2\pi$ to $\theta$ and arrive at different representations of the same complex number. For simplicity, we usually just stipulate that $0 \leq \theta < 2\pi$.
More generally, to take the $n$th roots of a complex number $z = re^{i\theta}$ with $r > 0$, multiply each $n$th root of $e^{i\theta}$ by the unique positive $n$th root of $r$, that is, $r^{1/n}$.
