Fundamental Theorem of Algebra/Proof 3

Theorem
Every non-constant polynomial with coefficients in $\C$ has a root in $\C$.

Proof
Let $p: \C \to \C$ be a complex, non-constant polynomial. We assume that $p \left({z}\right) \ne 0$ for all $z \in \C$.

Now consider the closed contour integral $\oint\limits_{\gamma_R} \tfrac{1}{ z \cdot p(z) } \,\mathrm dz $, where $\gamma_R$ is a circle with radius $R$ around the origin.

Since $\tfrac{1}{ z \cdot p(z) }$ is holomorphic in the whole complex plane with the exception of the origin, Cauchy's integral theorem implies that the value of this integral is independent of $R \ne 0$.

On the one hand, one can calculate the value of this integral in the limit $R \to 0$ (or use the Residue Theorem), using the substitution $z = R e^{i \phi}$:
 * $\lim\limits_{R \to 0} \oint\limits_{\gamma_R} \tfrac{1}{ z \cdot p(z) } \,\mathrm dz =

\tfrac{1}{p(0)} \lim\limits_{R \to 0} \oint\limits_0^{2 \pi} \tfrac{1}{R e^{i \phi}} \, i R e^{i\phi} \mathrm d\phi = \tfrac{2 \pi i}{p(0)} \ne 0.$

On the other hand, one can find an upper bound for the absolute value of the integral:
 * $\left|\oint\limits_{\gamma_R} \tfrac{1}{ z \cdot p(z) } \,\mathrm dz\right| \le

2 \pi R \max\limits_{|z| = R} \tfrac{1}{|z \cdot p(z)|} = 2 \pi \max\limits_{|z| = R} \tfrac{1}{|p(z)|}$ But this goes to zero for $R \to \infty$.

We arrive at a contradiction, and hence the assumption that $p \left({z}\right) \ne 0$ for all $z \in \C$ must be wrong.