TITLE: What do polynomials look like in the complex plane?
QUESTION [18 upvotes]: I have a hard time visualizing the fundamental theorem of algebra, which says that any polynomial has at least one zero, superficially I know this is true as every polynomial must have either an imaginary zero or real zero, but how do I visualize this in the complex plane?
For example if we have a real polynomial, we know that it is zero when it crosses the x axis this is because $y = 0$, however if $f(z) = 0$, then it must be the case that $f(z) = w = u+iv = 0+i0=0$ therefore every zero in $f(z)$ passes the origin? That does not make sense to me, what am I missing here?

REPLY [2 votes]: I read a part of a book called "Visual Complex Analysis" by Tristan Needham, and in it there was a visualization of complex polynomial. I highly recommend looking into that.
We know from the polar form of complex numbers that the product of complex numbers has the modulus equal to the product of the moduli of the factors and an argument equal to the sum of the arguments of the factors. So, if we have a polynomial of the form $f(z)=(z-c_{1})(z-c_{2})(z-c_{3})$ (for example), then we know that if we have a $z$ sufficiently close to a root ($c_{1}$,$c_{2}$, or $c_{3}$), then the modulus of $f(z)$ will also be small.