The formula for the $n$-th root of a complex number $r(\cos\delta+i\sin\delta)$ is:
$$r^{1/n}\left(\cos\frac{\delta+2k\pi}{n}+i\sin\frac{\delta+2k\pi}{n}\right)$$
where $k$ is an integer. Does this imply that complex numbers have infinitely-many $n$-th roots?
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Bear in mind that if $k = j*n + r$ where $r$ is the remainder of dividing $k$ by $n$ then $\sin (\frac {d + 2k\pi}n) = \sin (\frac {dn + 2r}n + 2j\pi) = \sin (\frac {dn + 2r}n$.
Same thing for $\cos (\frac {d + 2k\pi}n)$
So you only have as many solutions as possible remainders. And the remainders can be from $0$ to $n-1$. So there are $n$ solutions and it is finite. It's just that the values repeat after that.
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Note that $$ r^{1/n}\left(\cos\frac{\delta+2k\pi}{n}+i\sin\frac{\delta+2k\pi}{n}\right)$$ is periodic and it will only generate n solutions.
For example if you evaluate it at $k=0$ you get $$ r^{1/n}\left(\cos\frac{\delta}{n}+i\sin\frac{\delta}{n}\right)$$ Now if you evaluate it at k=n, you get $$ r^{1/n}\left(\cos\frac{\delta+2n\pi}{n}+i\sin\frac{\delta+2n\pi}{n}\right)$$
$$=r^{1/n}\left(\cos(\frac{\delta}{n}+2\pi)+i\sin (\frac{\delta}{n}+2\pi) \right)$$
$$= r^{1/n}\left(\cos\frac{\delta}{n}+i\sin\frac{\delta}{n}\right)$$
Remember that $\sin$ and $\cos$ are periodic functions. The way the n-th root is defined, while you can choose any integer value of $k$ to calculate the root you will find that if two values of $k$ differ by $n$, then they will result in the same root.
For example, the square roots of 1 are:
$\begin{eqnarray} (1+0i)^\frac{1}{2} & = & (1\mbox{ cis }0)^\frac{1}{2} \\ & = & (1)^\frac{1}{2} (\cos(0 + \pi k) + i\sin(0 + \pi k)\\ & = & \cos(\pi k) + i \sin(\pi k) \\ & = & \cos(0) + i \sin(0), \cos(\pi) + i \sin(\pi), \cos(2\pi) + i \sin(2\pi), \ldots \\ & = & 1 + 0i, -1 + 0i, 1 + 0i, -1 + 0i, \ldots \\ & = & 1, -1, 1, -1, \ldots \end{eqnarray}$
in other words, $1^\frac{1}{2} = \pm 1$, which we already knew. But it's the same for any number and any integral root - whenever $n | (k' - k)$, the $k'$th and $k$th $n$th roots will be the same value.
However, it does mean that you can have infinitely many distinct roots if $n$ is an irrational number, but that's a more complicated topic where you start having to pick what the "primary" root is.