Why do complex roots come in pairs? First time using this website so excuse me for using the body like this.
Complex roots of polynomials with real coefficients come in conjugate pairs because otherwise they wouldn't have real coefficients! Suppose a polynomial $p(z)=a_{0}+a_{1}z+ \ldots + a_{n}z^{n}$ has a complex root $\alpha$. Then $$\overline{p(\alpha)}=\overline{a_{0}+a_{1}\alpha+ \ldots + a_{n}\alpha^{n}}=\bar{a_{0}}+\bar{a_{1}}\bar{\alpha}+ \ldots + \bar{a_{n}}\bar{\alpha}^{n}\\=a_{0}+a_{1}\bar{\alpha}+ \ldots + a_{n}\bar{\alpha}^{n}=\bar{0}=0$$
So $\bar{\alpha}$ is also a root.
Because, in the complex plane, a root is also a rotation. In general, think of a complex number in polar representation:
$$z=r\cdot e^{i\phi}$$ Then taking the square root gives you:
$$\sqrt{z} = z^{-1/2} = \sqrt r\cdot e^{i\phi/2}$$
But - the angle $(\phi/2+\pi)$ also satisfies the demands for being half the angle $\phi$, because:
$$2\cdot(\phi/2+\pi) = \phi+2\pi = \phi$$

It all stems from the fact that rotations live in on a circle, which is a cyclic space (obviously). When you do arithmetic on angles you must remember that a $full\ rotation = no\ ratation$, and from that it follows that when solving simple equations like $2x=y$ you end up with more than a single solution.