A new very short proof of the fundamental theorem of algebra

I've always been intrigued by the fundamental theorem of algebra (every nonconstant polynomial with complex coefficients has a root), not least because I don't know any proof which uses algebra only. Earlier, I posted an easy proof in this blog, one that uses Cauchy's theorem.There is a recent proof (Oswaldo Rio Branco de Oliveira, Mathem. Intellig., March 2011), which is almost trivial. It goes as follows (and this is a chance for me to see if the embedded LaTeX script works...):

Let $P(z)$ be a polynomial of degree $n$. Since $|P(z)|$ is a nonnegative continuous function, tending to $\infty$ as $|z|$ tends to $\infty$, it has a minimum at some point $z_0$:

$|P(z)| \ge |P(z_0)|$, for all $z$.

By division of $P(z)-P(z_0)$ by $z-z_0$, write

$P(z) = P(z_0) + (z-z_0)^k Q(z-z_0),$

where $Q(0) \not = 0$. Since $P(z)$ is nonconstant, the integer $k$ is $\ge 1$.
Therefore

$|P(z_0) + (z-z_0)^k Q(z-z_0)|^2 \ge |P(z_0)|^2$, for all $z$,

and, expanding the square,

$|z-z_0|^{2k} |Q(z-z_0)|^2 + 2 \Re \{ (z-z_0)^k Q(z-z_0) \overline{P(z_0)}\} \ge 0$, for all $z$.

Let $z=z_0 + r e^{i \theta}$, divide by $r^k$, and let $r$ tend to $0$. We obtain

$\Re \{ e^{i k \theta}  Q(0) \overline{P(z_0)}\} \ge 0$,  for all real $\theta$.

It is an easy exercise in algebra that, if $\alpha$ is a complex number such that $\Re \{ e^{i k \theta} \alpha\} \ge 0$ for all $\theta$, then $\alpha=0$. Hence $Q(0) \overline{P(z_0)}=0$. Since $Q(0) \neq 0$, we obtain $P(z_0)=0$.