# Counting roots of polynomial inside $S^1$

I would like to ask for a hint to this problem:

Let $p$ a polynomial function on $C$ with no root on $S^1$. Show that the number of roots of $p$ with $|z|<1$ is the degree of the map $q: S^1 \to S^1$ given by $q(z)=p(z)/|p(z)|$.

This problem appears in Peter May's book, concise course in Algebraic Topology. Assuming the fundamental theorem of algebra, my idea so far is to collapse every zero inside $S^1$ through a homotopy to zero, explicitly, if $a_1, ..., a_k$ are the roots inside $S^1$ take $h(z,t)=(z-t*a_1)...(z-t*a_k)*p_2(z)$, where $p_2(z)$ is the part of the polynomial that has the zeros outside $S^1$, and consider $h(z,t)/|h(z,t)|$. Then, I end up with something like $z^k*p_2(z)/|p_2(z)|$. Next, I think of vanishing, somehow, the part involving $p_2$, but don't know how.

Any advice would be appreciated.

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You can use the homotopy $z^kp_2(tz)/|p_2(tz)|$ which is well defined because $p_2$ has no zeros in the unit disk – omar Feb 18 '14 at 14:32
Hey, omar thanks, it was so easy. I'm a little bit ashamed I did not see that. By the way, do you know of any other way to solve this? – Juan Feb 18 '14 at 16:12

Here is an idea (caution, I did not try this but it is what I would try if I was attempting the question):

The degree of a continuous map $S^n \to S^n$ is called the winding number if $n=1$ and it may be computed using the following formula:

$$\text{ winding number} = {1 \over 2 \pi i}\oint_C {dz \over z}$$ where $C$ is a curve around $0$. In your question $C = q(z)$ and therefore the winding number of $q$ becomes

$$\text{ winding number of q} = {1 \over 2 \pi i}\oint_{q \circ \gamma} {dz \over z}$$

where $\gamma(t) = e^{2 \pi i t}$ and $t \in [0,1]$.

So now we have the degree of $q$. It remains for us to count the roots of $p$. To this end, note that if $p$ has a root at $z_0$ then ${1 \over p}$ has a pole at $z_0$. To count the poles of a map inside a closed curve you may use the residue theorem:

Let $\beta$ be any curve inside $S^1$ and outside all roots of $p$. Then by the residue theorem

$$\oint_\beta {1 \over z} dz = 2\pi i \sum_{k=1}^n \mathrm{Res}({1\over p}, a_k)$$

where $a_k$ are the roots of $p$. Now all that remains to be done is to calculate the residues and then show that the two are equal.

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It's not the residue of $\frac{1}{p}$ that you want. On the one hand, that residue changes when we multiply $p$ with a (nonzero) constant - and neither the degree nor the number of zeros in the unit disk does - on the other, when $p$ has a multiple zero, the residue may be $0$. What you want is the residue of $\frac{p'}{p}$, since that counts the order of the zero: $p(z)=(z-a)^kq(z)$ with $q(a)\neq0$ leads to $$\frac{p'(z)}{p(z)}=\frac{k(z-a)^{k-1}q(z)+(z-a)^kq'(z)}{(z-a)^kq(z)}=\frac{k}{‌​z-a}+\frac{q'(z)}{q(z)}.$$ Conveniently, the integral $$\frac{1}{2\pi i}\int_{S^1}\frac{p'(z)}{p(z)}\,dz$$ – Daniel Fischer Feb 22 '14 at 16:35
not only counts the zeros of $p$ (minus the poles of $p$, but there are none of those) inside the unit disk, it also is the winding number of $p\circ \kappa$, where $\kappa(t) = e^{it}$ around $0$, i.e. the degree of $p\lvert_{S^1}$. – Daniel Fischer Feb 22 '14 at 16:37
@DanielFischer That $q$ in your first comment is not the same $q$ as in the question, right? – Rudy the Reindeer Feb 24 '14 at 15:59
Yes, it isn't. Sorry for confusing notation. Replace that with $g$ or whatever else wasn't used. The point is that if a holomorphic function $f$ has a zero of order $k$ in $a$, we can write it as $f(z) = (z-a)^k\cdot g(z)$ with a holomorphic $g$ that has $g(a)\neq 0$. – Daniel Fischer Feb 24 '14 at 16:19
@DanielFischer Yes, thanks. I was 99% sure it wasn't but... you (=I) never know. : ) I think I understand: It looks like if $a_n$ are the zeros of $p$ then the fraction ${p' \over p}$ expands as a sum $\sum_n {k_n \over z-a}$. Computing the integral of the sum yields $\sum_n k_n$ because $\oint {1 \over z-a}dz$ around $a$ is $1$. It's pretty neat. – Rudy the Reindeer Feb 24 '14 at 16:26