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
 * $\displaystyle \oint \limits_{\gamma_R} \frac 1 {z \cdot p \left({z}\right)} \, \mathrm d z$

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

Since $z$ and $p(z)$ are holomorphic and $p(z)$ is assumed to have no zeroes, $\dfrac 1 {z \cdot p \left({z}\right)}$ is holomorphic in the whole complex plane with the exception of the origin (because the product and the quotient of holomorphic functions is again holomorphic for all $z \in \C$ where the quotient is defined). Hence 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 parametrization $z = R e^{i \phi}$ 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 arrive at a contradiction, and hence the assumption that $p \left({z}\right) \ne 0$ for all $z \in \C$ must be wrong.