# Winding number of a polynomial

Consider $f(z) = c_n z^n + ... + c_1 z + c_0$, where $c_n\ne 0$. Let $C_R$ be the circle of radius $R$ centred at the origin, oriented counterclockwise. Prove that the winding number of $f\circ C_R =n$ for $R$ sufficiently large.

My approach:

Parametrize $C_R$ as $\gamma(t) = Re^{it}$. Then $$\frac{1}{2\pi i}\int_{C_R} \frac{f'(z)}{f(z)}dz=\frac{1}{2\pi i}\int\limits_0^{2\pi} \frac{f'(\gamma(t))\gamma'(t)}{f(\gamma(t))}dt$$ $$=\frac{1}{2\pi }\int\limits_0^{2\pi} \frac{nc_n (Re^{it})^n + ... + c_1 Re^{it}}{c_n (Re^{it})^n+...+c_1Re^{it}+c_0}dt$$

I thought I got stuck here, but now I'm thinking: maybe I should take the limit as $R\to \infty$ of the integral above, take the limit in the integral (since the limit is not in terms of $t$), and then observe that the integrand becomes $ndt$, and so the integral comes to $\frac{2\pi n}{2\pi}=n$? Would this approach be correct? I think so, because $R$ should go to infinity in order to encompass all possibilities for all zeros of $f(z)$.

• Correct. You can also note that asymptotically (for large $R$), the winding number will be that of the dominant term, which is $z^n$, i.e. a rotation of winding $n$. Jul 18 '16 at 9:50
• That looks fine, but: haven't you studied yet the (Cauchy's) Argument Principle? Because if you have then the proof, if I'm not mistake, takes one line at most and looks much less messy. Jul 18 '16 at 10:00
• @DonAntonio By the Argument Principle, we have to "already" know that f has n roots. But this proof is another proof of the Fundamental Theorem of Algebra, where we assume we don't yet know that f has n roots, and deduce it in the end, by the Argument Principle. Jul 18 '16 at 14:56
• @sequence That wasn't clear from the beginning, and you also wrote "$\,R\;$ should go to infinity to encompass all possibilities for all zeros of $\;f(z)\;$", so I didn't even think of the FTA. Jul 18 '16 at 17:12
• @DonAntonio I think I wasn't clear enough in my original post. When we let $R\to \infty$ we want to account for all possibilities of "covering" any zeros of $f$, since $f$ is a general polynomial. That's what I should've said. Jul 18 '16 at 17:20

By the Argument Principle, directly we get for $\;R\;$ big enough so that all the roots of the polynomial are within the circle $\;|z|=R\;$:
$$\frac1{2\pi i}\oint_{C_R}\frac{f'(z)}{f(z)}dz=n$$
• By the Argument Principle, we have to "already" know that $f$ has $n$ roots. But this proof is another proof of the Fundamental Theorem of Algebra, where we assume we don't yet know that $f$ has $n$ roots, and deduce it in the end, by the Argument Principle. Jul 18 '16 at 14:56
Suppose wlog $c_n =1$. $$\int_{|z|=R}\frac{f'(z)}{f(z)}\,dz = \int_{|z|=R}\left(\frac{f'(z)}{f(z)} - \frac{n}z + \frac{n}z\right)dz = \int_{|z|=R}\left(\frac{f'(z)}{f(z)} - \frac{n}z\right)dz + 2\pi i n$$ and for some $M>0$ $$\left|\int_{|z|=R}\left(\frac{f'(z)}{f(z)} - \frac{n}z\right)dz\right| = \left|\int_{|z|=R}\frac{zf'(z) - nf(z)}{zf(z)}\,dz\right|\le\frac{MR^{n-1}}{R^{n+1}}2\pi R = \frac{2\pi M}R.$$