Winding number of a polynomial

Question

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 $f(z)$.

Answer 1

By the Argument Principle, directly we get for $\;R\;$ big enough so that all the roots of the polynomial are within the circle $\;|z|=R\;$:

$$\frac1{2\pi i}\oint_{C_R}\frac{f'(z)}{f(z)}dz=n$$

and we're done.

Answer 2

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 $M>0$ $$ \left|\int_{|z|=R}\left(\frac{f'(z)}{f(z)} - \frac{n}z\right)dz\right| = \left|\int_{|z|=R}\frac{zf'(z) - nf(z)}{zf(z)}\,dz\right|\le\frac{MR^{n-1}}{R^{n+1}}2\pi R = \frac{2\pi M}R. $$