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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 $f(z)$.

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  • $\begingroup$ 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 $n$. $\endgroup$ Jul 18 '16 at 9:50
  • $\begingroup$ 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. $\endgroup$
    – DonAntonio
    Jul 18 '16 at 10:00
  • $\begingroup$ @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. $\endgroup$
    – sequence
    Jul 18 '16 at 14:56
  • $\begingroup$ @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. $\endgroup$
    – DonAntonio
    Jul 18 '16 at 17:12
  • $\begingroup$ @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. $\endgroup$
    – sequence
    Jul 18 '16 at 17:20
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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.

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  • $\begingroup$ 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. $\endgroup$
    – sequence
    Jul 18 '16 at 14:56
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    $\begingroup$ @sequence, you should have told us in the question that you were trying to prove the FTA. $\endgroup$ Jul 19 '16 at 7:35
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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 $M>0$ $$ \left|\int_{|z|=R}\left(\frac{f'(z)}{f(z)} - \frac{n}z\right)dz\right| = \left|\int_{|z|=R}\frac{zf'(z) - nf(z)}{zf(z)}\,dz\right|\le\frac{MR^{n-1}}{R^{n+1}}2\pi R = \frac{2\pi M}R. $$

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