Use the minimum modulus principle (suppose $f$ is a non-constant analytic function in a domain $D$, and that $a$ $\in$ $D$. a is not a local minimum point of $f$ unless $f(a)=0$). Use this fact to prove the fundamental theorem of algebra.
Here is my attempt:
Proof: Assume that $p(z)=a_0+a_1z+…….+a_nz^n$ for $n \geq 1$. Assume also that $p(z) \neq 0$ and that $p(z)$ has no root. Observe also that $p(z)$ is nonconstant. Now, let $f(z)=\frac{1}{p(z)}$. As $p(z)$ is analytic (because all polynomials are analytic) and as $p(z)$ is nonconstant, we have that $f(z)$ is a nonconstant analytic function on the domain $D$. Now let $a$ denote the local minimum point. We then have $f(a)=0$. Thus $f(a)=\frac{1}{p(a)}=0$. But this is a contradiction as $p(z) \neq 0$ for all $z \in \mathbb{C}$.
Is this right, or on the right track? I have a feeling I made some leaps in logic.
Best Answer
You don't use explicitly the fact that $p$ is a polynomial function. Your reasoning seems work with the exponential function...
You have to prove that $p$ admits a local minimum, which requires using the fact that $p$ is algebraic.
EDIT: In fact you can prove that $p$ reaches a global minimum.