The theorem in question is:
If $p(z) = z^n + a_1z^{n-1} + \cdots + a_n$, then either $p$ is constant or $p$ has a root in $\mathbb C$.
The following is a topological proof using the fundamental group of the $1$-sphere $S^1$:
Assume $p$ has no roots. Let $$g_n(s) = \frac{p(re^{2\pi is}/p(r)}{|p(re^{2\pi is}/p(r)|}.$$ For each $r\geqslant0$, this defines a loop in $S^1\subset\mathbb C$ based at $1$. They are all homotopic and $g_0$ is the constant loop. So $[g_r]=[\omega_o]$ in $\pi_1(S^1)$. Let $R>\max\{1,|a_1+\cdots+a_n|\}$. If $|z|=R$, then
$$
|z^n| = R^n = RR^{n-1}>(a_1z^{n-1}+\cdots+a_n)R^{n-1}\geqslant |a_1z^{n-1}+\cdots+a_n|.$$ Hence $p_t(z) = z^n + t(a_1z^{n-1}+\cdots+a_n)$ does not have any roots with $|z|=R$ for $t\in[0,1]$. Consider
$$
g_R(t,s) = \frac{p_t(re^{2\pi is}/p_t(r)}{|p_t(re^{2\pi is}/p_t(r)|}.
$$
This is a homotopy between $[g_R]$ and $[g_R(0,\cdot)] = [\omega_n]$. Hence $[\omega_0] = [\omega_n]$, so $n=0$.
I do not see how we conclude from $n=0$ that $p$ is constant. Why is this the case?
Best Answer
What does a polynomial of degree zero look like? Can you write down two or three of them?