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 $P(z_0)=0$.

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