We know the polynomial of odd degree must have a real root.
Because of the the intermediate value theorem:
Proof: < 0 for some x
> 0 for some x
+ IVT
2 things to point out:
i)not necessarily true for even degree polynomials (for example, $x^2 +1 = 0$ has no real root)
ii)not necessarily true for rational numbers. $x^3 +2=0$ has no rational root.
Suppose $f(z) = a_0 + a_1z + a_2z^2 + a_3z^3 + a_4z^4+ ... + a_nz^n$
Is there a $z_0$ element of C such that $f(z_0)=0$?
That is, does every complex polynomial have a root in C?
Question: What does Newton's method do for $x^2+1$ starting at $x_0=1?$
Theorem: every f(z) above has n roots of C.
Let's assume a_n=1 (divide out by it otherwise)
1. Prove that each has a single root
2. Pull it out, by deduction, it will have n1 single roots.
so we can write $f(z)=(zr_1)(zr_2)...(zr_n)$ $r_j$element of C
if $f(r_1)=0$, then we can write $f(z)=(zr_1)q(z)+c$
and then $f(r_1)=0 => (r_1r_1)q(r_1) +c = 0$
c = 0
so since q(z) has degree n1, we can use induction hypothesis to get $q(z) = (zr_2)...(2r_n)$
so it is enough to show that $f(z)=0$ has at least one root!
$f(z) = a_0 + a_1z + a_2z^2 + a_3z^3 + a_4z^4+ ... + a_nz^n$
if a_0 = 0, then z=0 is a root. And we would be done.
So we may assume that a_0 does not equal 0.

$f(z) = z^n ( 1 + a_{n1}/z + a_{n2}/z^2 + ...)$
$f(z) = z^n$
$z=Re^{i*theta}$
where R is huge
g(z) = z^n which is n copies of a circle of radius R^n
if we change our radius continuously from E (not containing 0) to R (containing 0) at some point we must change from not containing to containing the origin. Thus there must be a value r_0 and a rho =theta_0 so
$f(r_0e^{i*theta_0})=0$ so $z=r_0e^{i*theta_0}$ is a root.
[x == 1/2*(1/2*sqrt(5)  1/2)^(1/3)*(I*sqrt(3) + 1) + 1/2*(I*sqrt(3) + 1)/(1/2*sqrt(5)  1/2)^(1/3), x == 1/2*(1/2*sqrt(5)  1/2)^(1/3)*(I*sqrt(3) + 1) + 1/2*(I*sqrt(3) + 1)/(1/2*sqrt(5)  1/2)^(1/3), x == (1/2*sqrt(5)  1/2)^(1/3)  1/(1/2*sqrt(5)  1/2)^(1/3)] [x == 1/2*(1/2*sqrt(5)  1/2)^(1/3)*(I*sqrt(3) + 1) + 1/2*(I*sqrt(3) + 1)/(1/2*sqrt(5)  1/2)^(1/3), x == 1/2*(1/2*sqrt(5)  1/2)^(1/3)*(I*sqrt(3) + 1) + 1/2*(I*sqrt(3) + 1)/(1/2*sqrt(5)  1/2)^(1/3), x == (1/2*sqrt(5)  1/2)^(1/3)  1/(1/2*sqrt(5)  1/2)^(1/3)] 
