Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Thanks @ZevChonoles, I should really work on my copy-editing... Thank you.. (: –  1LiterTears Jul 21 '13 at 19:53
    
Maybe it would be helpful to consider the function $\frac1{2\pi i}\log\frac{p(z)}{|p(z)|}$. –  Kirill Jul 21 '13 at 20:02
    
Hi @Kirill, I put in some thought but didn't get it. 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. Could you clear it up? thanks X-) –  1LiterTears Jul 21 '13 at 20:22
add comment

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.