This is problem 8-5 in Lee's Topological Manifolds:
Prove that every nonconstant polynomial on one complex variable has a zero. [Hint: if $p(z)=z^n+a_{n-1z^{n-1}}+ \dots + a_0$, write $p_\epsilon(z)=\epsilon^np(z/\epsilon)$and show that there exists $\epsilon >0$ such that $|p_\epsilon(z) -z^n|<1$ when $z\in\mathbb{S}^1$. Prove that if $p$ has no zeros, then $p_\epsilon|_{\mathbb{S}^1}$ is homotopic to $p_n(z)=z^n$, and use degree theory to derive a contradiction.]
This is what I got so far:
$|p_\epsilon(z) -z^n|=|a_{n-1}z^{n-1}\epsilon+ \dots + a_0\epsilon^n|$ if $z\in\mathbb{S}^1$ then $$|a_{n-1}z^{n-1}\epsilon+ \dots + a_0\epsilon^n|\leq \epsilon|a_{n-1}|+\dots+ \epsilon^n|a_0|$$ since the right side goes to $0$ as $\epsilon \to0$, we can definitively find an $\epsilon>0$ such that $|p_\epsilon(z) -z^n|<1$ when $z$ is in the unit circle. But, I don't know how to use this hint.
I also know this: since $p$ has no zeros, $p_\epsilon$ has no zeros either and $$\frac{p_\epsilon|_{\mathbb{S}^1}}{|p_\epsilon|_{\mathbb{S}^1}|}:\mathbb{S^1} \to\mathbb{S^1}$$ is a continuous function that has an extension to the unit disk (namely $p_\epsilon$) and therefore has degree $0$. If I can show that this function is homotopic to $p^n$, I would get $n=0$, which means the polynomial was constant and done! But I have no idea what to do with the hinted inequality. My understanding of $p_\epsilon$ is that it is the same polynomial $p$ scaled smaller? Also, $p_\epsilon$ seems to be an homotopy between $p$ and $p^n$ but $\epsilon >0$.
As I wrote in the comments, the point is that two maps on the circle that differ by at most $1$ are homotopic, if one of them has constant modulus $1$.
Indeed let $f,g: S^1\to \mathbb{C}^*$ be such that $|f(x)-g(x)|<1$ for all $x$, and $|g(x)|=1$ for all $x$. Then consider $H(x,t) = tf(x) + (1-t)g(x)$.
Clearly this is continuous $S^1\times I \to \mathbb{C}$. But now assume $H(x,t) = 0$ for some $x,t$. Then $tf(x) + (1-t)g(x) = 0$, then $tf(x) = (t-1)g(x)$, so $f(x) = \frac{t-1}{t}g(x)$ (because $t$ can't be $0$), so $f(x) - g(x) = \frac{-1}{t}g(x)$, and so $|f(x)-g(x)| = \frac{1}{t}\geq 1$ which is absurd.
Thus $H:S^1\times I \to \mathbb{C}^*$ is continuous.
In particular $p_\epsilon$ and $p^n$ are homotopic, and $p$ is homotopic to $p_\epsilon$ through $K(x,t) = p_{\gamma (t)}(x)$ (which lands in $\mathbb{C}^*$ because $p$ has no roots) where $\gamma$ is any continuous path in $\mathbb{R}_+^*$ from $1$ to $\epsilon$