Fundamental Theorem of algebra using cauchy integral formula

Question

Is there any way we can prove the fundamental theorem of algebra by applying the cauchy integral formula to the function $ f(z)=\dfrac{1}{zp(z)}$? To prove this, I've tried sort of emulating the proof of the fundamental theorem of algebra using Liouvilles theorem by making use of cauchy integral formula involving the second derivative, but it isnt seeming to work well, with the point z=0 posing as a problem point making me unable to expand my circle radius. Any help would be appreciated

Answer 1

Suppose $p$ is a non-constant polynomial with no zeros. It is esy to see that $|p(z)| \to \infty$ as $|z| \to \infty$. Hence $\int_{|z|=R} \frac 1 {zp(z)}dz=i\int_0^{2\pi} \frac 1 {p(Re^{i\theta})} d\theta \to 0$ as $ R \to \infty$. But the integral equals $\frac {2\pi i} {p(0)}$ for every $R>0$ and we have a contradiction.