Number of roots the degree of the map? Let $p$ be a polynomial function on $\mathbb{C}$ which has no root on $S^1$. My question is as follows: does the number of roots, up to multiplicity, of $p(z) = 0$ with $|z| < 1$ necessarily equal the degree of the map $\widehat{p}: S^1 \to S^1$ specified by $\widehat{p}(z) = p(z)/|p(z)|$?
 A: A simpler solution than the one of Lee Mosher is as follows. Let $z_1, \dots, z_n$ be the roots of $p$ inside the unit circle, and let $w_1, \dots, w_m$ be the roots of $p$ outside the unit circle. So$$p(z) = (z - z_1) \dots (z - z_n) (z - w_1) \dots (z - w_m).$$Define a homotopy from $\alpha q$ to $\widehat{p}$ where $\alpha \in S^1$ and $q(z) = z^n$:$$H(z, t) = {{(z - tz_1) \dots (z - tz_n) (tz - w_1) \dots (tz - w_m)}\over{|(z - tz_1) \dots (z - tz_n)(tz - w_1) \dots (tz - w_m)|}}.$$For all $t$, we have that $tz_i$ is inside the unit circle for $i = 1, \dots, n$, and $w_j/t$ is outside the unit circle for $j = 1, \dots, m$. Thus, $H$ is continuous and is the desired homotopy, with$$\alpha = {{(-w_1)\dots(-w_m)}\over{|w_1 \dots w_m|}}.$$Let $\gamma(t) = e^{2\pi it}$, a representative of the generator $\pi_1(S^1)$. Then $q(\gamma)(t) = e^{2\pi tin}$, i.e., $q_*([\gamma]) = [\gamma]^n$. So $\deg q = n$. Suppose $\alpha = e^{2\pi i \rho}$. Then multiplication by $\alpha$ is homotopic to the identity by$$H(z, t) = ze^{2\pi i(1 - t)\rho}$$and so has degree $1$. Composition of maps induces multiplication of degrees, so$$\deg \widehat{p} = \deg (\alpha q) = n.$$
A: That is almost correct: if you count the roots with multiplicity it is exactly correct. The proof is a homology argument. 
Let $z_1,..,z_K$ be the rools of modulus $<1$. 
Let $D_1,…,D_K$ be small round disks centered on $z_1,…,z_K$ respectively, chosen to be pairwise disjoint and disjoint from $S^1$. 
Let $F = D^2 - (\text{int}(D_1) \cup \cdots \cup \text{int}(D_K)$. 
The map $p \bigm| F$ may be regarded as a 2-cycle in $\mathbb{C}-0$, and as such it is a homology in $\mathbb{C}-0$ from the 1-cycle $p \bigm| S^1$ to the 1-cycle $p \bigm| (\partial D_1 \cup \cdots \cup \partial D_K)$. Those two 1-cycles therefore represent the same element of $H_1(\mathbb{C}-0;\mathbb{Z}) \approx \mathbb{Z}$. The 1-cycle $p \bigm| S^1$ is homotopic to $\hat p$ and therefore represents the degree of $\hat p$. The 1-cycle $p \bigm| (\partial D_1 \cup \cdots \cup \partial D_K)$ is the sum of the individual 1-cycles $p \bigm| \partial D_k$, and the latter represents the multiplicity of the zero at $z_k$.
