Intuitive explanation as to why the complex numbers on a complex plane split when taking the root? When taking the root of a complex number, it splits into several points on the complex plane. I can see how this comes about mathematically, but this is hard for me to understand intuitively. 
Is this just a convenient coincidence?

 A: The rule for multiplication of complex numbers is to add the arguments and multiply the length. That is, multiplying by a complex number corresponds to a rotation combined with a scaling operation. Thus, if you want to multiply a complex number $n$ times by itself and produce a given modulus $r$ and a given angle $\theta$, the modulus has to be $\sqrt[n] r$, but there are actually $n$ different angles such that rotating by them $n$ times produces a rotation of $\theta$ radians.
But really, I'm not sure there's even anything to explain. If you have a bunch of things that you can multiply together, then an "$n$-th root" just means "something that when multiplied by itself $n$ times produces a given result". Why, intuitively, would there only be one of those?
A: It may be easier to first rationalize that the complex $n^{th}$ roots of unity are the vertices of a regular $n$-gon inscribed in the unit circle.
Then, let $\omega$ be a primitive $n^{th}$ root of unity so that $\omega^n=1$. If $z_0$ is any of the $n^{th}$ roots of $z$, it follows that $w^k z_0$ is also an $n^{th}$ root of $z$, since $(\omega^kz_0)^n=(\omega^{n})^kz_0^n = 1^k \cdot z = z$. Therefore the complete set of the $n$ $n^{th}$ roots of $z$ is $\{z_0, \omega z_0, \omega^2 z_0, \dots, \omega^{n-1}z_0\}$ which are the vertices of a regular $n$-gon centered on the origin and "anchored" at $z_0\,$. That $n$-gon is just the regular $n$-gon of the $n^{th}$ roots of unity $\{1,\omega,\omega^2,\dots,\omega^{n-1}\}$ scaled by $|z_0|=\sqrt[n]{|z|}$ and rotated by $\arg(z_0)\,$.
A: I am not sure what you mean by understanding it intuitively..
Consider $z = r(\cos\theta + i\sin\theta)$ then the roots are 
$$ z_k = r^{\frac{1}{n}}(\cos\frac{\theta + 2\pi k}{n} + i\sin\frac{\theta + 2\pi k}{n}),\ \ k = 0, \ 1,\ldots,\ n-1 $$
This always gives $n$ different roots. Maybe this is what you meant that you understood it mathematically..
Geometrically, when you multiply $2$ complex numbers, $z_1=r_1e^{i\theta_1},\ z_2=r_2e^{i\theta_2}$, you get $z=r_1r_2e^{i(\theta_1+\theta_2)}$
This effectively takes $z_1$ and multiplies its radius by that of $z_2$ and rotates it by the angle of $z_2$.  
Now going back to the roots $z_k$ of $z$, their radius is $r^{\frac{1}{n}}$, so multiplying them by themselves gives radius $r$, and their angle is $\frac{\theta + 2\pi k}{n}$ so rotating it $n$ times by the same angle gives $\theta + 2\pi k$
Does that make sense?
A: For complex numbers, when you say to take the "root", you are referring to all solutions to the following problem:
$$x^n=z$$
where $x$ is the $n$th root of $z$.  As a simple case, consider $n=2$ and $z=1$.  Then,
$$1^2=(-1)^2=1$$
$$x=\begin{cases}+1\\-1\end{cases}$$
Similar such happen for higher $n$, and quite a symetric splitting manner.
