Fundamental Theorem of Algebra/Proof 3

Proof
Let $p: \C \to \C$ be a complex, non-constant polynomial.

that $\map p z \ne 0$ for all $z \in \C$.

Now consider the closed contour integral:
 * $\ds \oint \limits_{\gamma_R} \frac 1 {z \cdot \map p z} \rd z$

where $\gamma_R$ is a circle with radius $R$ around the origin.

By Derivative of Complex Polynomial, the polynomial $z \cdot \map p z$ is holomorphic.

Since $\map p z$ is assumed to have no zeros, the only zero of $z \cdot \map p z$ is $0 \in \C$.

Therefore by Reciprocal of Holomorphic Function $\dfrac 1 {z \cdot \map p z}$ is holomorphic in $\C \setminus \set 0$.

Hence the Cauchy-Goursat Theorem implies that the value of this integral is independent of $R > 0$.

On the one hand, one can calculate the value of this integral in the limit $R \to 0$ (or use Cauchy's Residue Theorem), using the parameterization $z = R e^{i \phi}$ of $\gamma_R$:

which is non-zero.

On the other hand, we have the following upper bound for the absolute value of the integral:

But this goes to zero for $R \to \infty$.

We have arrived at a contradiction.

Hence by Proof by Contradiction the assumption that $\map p z \ne 0$ for all $z \in \C$ must be wrong.