I can tell you that the Fundamental Theorem of Algebra states that each nth degree polynomial has exactly n roots.
If you think of the graph of a polynomial, the roots are where it crosses the x-axis (the "zeroes"). In the case where, let's say, a parabola doesn't cross at all, we still need to have two roots according to the F.T.A..
This is where complex analysis comes in, because the roots will no longer be real. For n=2 you can look at this case:
x^2+1=0
Will have solutions:
x^2=-1
x=\pm i
As you can see the solutions are complex conjugates of each other. This gives the symmetry in the complex plane. For an n^{th} degree polynomial, if can have two cases:
n\in \mathbb{Z} \text{ even}
n\in \mathbb{Z} \text{ odd}
If n is even, there can be at most \frac{n}{2} sets of imaginary roots. If n is odd there can be at most \frac{n-1}{2} sets of imaginary roots because the last root will not have a conjugate, and therefore must be real.
Another good example of this is when:
x^4+1=0
x=(-1)^{1/4}
So there must be exactly 4 roots since it is a 4th degree polynomial. Using a bit of complex analysis:
e^{i\pi}=-1
x=(e^{i\pi})^{1/4}
Now in complex analysis you will learn eventually that you can't just take the 1/4 power of complex numbers without considering the fact that they have multi-value properties, so in this case:
x=\exp{[i(\frac{\pi}{4}+\frac{2n\pi}{4})]}
Since every 2n\pi gives the same value (a full rotation in the circle), we must make sure we divide this by 4 as well to get the correct last answer. Now n ranges from 0 to 3. If you plot these points in the complex plane they will form a circle of radius 1. You can use deMoivre's formula to plot them:
\exp{i\theta}=\cos{\theta}+i\sin{\theta}
Does this make sense? I hope I didn't go over your head, I'm not sure what Calculus AB is. To plot using deMoivre's formula, just consider \cos{\theta} as the x-coordinate and \sin{\theta} as the y-coordinate.
