Why the degree of $p/|p|$ is zero on $\partial W^\prime?$

Question

Let $W$ be a smooth compact region in $\mathbb{C}$ whose boundary con­tains no zeros of the polynomial $p$. $p$ has only finitely many roots, $z_0, \dots, z_n$ in $W$. Around each $z_i$, circumscribe a small closed disk $D_i$, making the disks all disjoint from each other and from the boundary. Then $p/|p|$ is defined on $$W^\prime = W - \bigcup_{i=0}^n D_i.$$ Thus, $$\partial W^\prime = \partial W - \bigcup_{i=0}^n \partial D_i.$$

Then why the degree of $p/|p|$ is zero on $\partial W^\prime?$

One thought...

$$\frac{p(z)}{|p(z)|}\quad \text{and}\quad \frac{z^m}{|z^m|}=\left(\frac{z}{r}\right)^m$$ are homotopic maps of $S \to S^1$. Thus $p/|p|$ must have the same degree as $(z/r)^m$ - namely, $m$. So I find trouble concluding degree is $0$ from the boundary condition.

A second thought...

Should I use the fact that $\deg\Big(\frac{p(z)}{|p(z)|}\Big) = I\Big(\frac{p(z)}{|p(z)|},\{y\}\Big), y \in \partial W^\prime$, the I am not sure how to show $\Big(\frac{p(z)}{|p(z)|}\Big)^{-1}(\{y\}) = 0.$ In fact what I got is not necessarily equal to 0, consider the preimage of a point on the unit circle, whose preimage may be multiple points.

Answer 1

The map $p/|p|$ is a smooth map $W'\to S^1$, hence by the Boundary Theorem its degree on the boundary is $0$. This argument works in any dimension.