Fundamental Theorem of Algebra for highschool

Question

My teacher has told me about the Fundamental Theorem of Algebra, but I can't seem to find any proofs on it which I can understand. For something so important I'm hoping to find a proof that a highschool student can understand.

References are welcome too, thanks!

Answer 1

There is one intuitive proof for this theorem. Let $p(z)=\sum_{i=0}^na_iz^i$ where $a_i\in \mathbb{C}$. We need to show that $p$ has a zero in $\mathbb{C}$. Take a very large real number $R$ and consider the circle $|z|=R$ in $\mathbb{C}$. Since $R$ is large, plugging these $z$ into $p$, we have that $p(z)$ is some very large contour that is somewhat centered at $0$. (Since $R$ is large, the contribution of $a_n z^n$ dominates, so $p(z)\sim a_nz^n$). Similarly, consider the circle $|z|=r$ for some real number $r$ close to zero, then $p(z)\sim a_0$, hence $p(z)$ is some small contour near $a_0$.

When we let $r$ vary from very small to very large, then we get growing contours in $\mathbb{C}$ whose centers vary from $a_0$ to $0$. At some point some contour must go through the origin.

One can make an argument like this precise, however it is very difficult to do so.

Answer 2

Here's a proof that I've always liked.

Suppose $\forall z : P(z)\neq 0$ Let $$f(z) := |P(z)|$$ Then $f(z)$ is larger than some number $m>0$ and attains its minimum $m$ on some point $z_0$ (which must be finite because $P(z)$ goes to $\infty$ as $z$ goes to $\infty$)

Let $$Q(z):= P(z-z_0)$$ We have $Q(0) = a$ s.t $|a| = m$ and $|Q(z)|\geq m$

Then $$Q(z) = a+ \sum_1^n a_k z^k \underset{z\to 0}{=} a + a_p z^p + o(z^p)$$ Where $p$ is the first number for which $a_p\neq 0$

So $$|Q(z)| \underset{z\to 0}{=} |a+a_pz^p| + o(z^p)$$ If we choose $a_p z^p = aw^p$ ,i.e $z = \left(\frac{a}{a_p}\right)^{1/p}w$,
we get $$Q(z) \underset{w\to 0}{=} |a||1+w^p| + o(w^p) = m|1+w^p| + o(w^p)$$ So if we take $w = (-t)^{1/p}$ s.t $t$ is a positive real number. $$Q(z) \underset{t\to 0}{=} m|1-t| + o(t) = m(1-t) + o(t) $$ Which is strictly smaller than $m$ for small enough $t$.

Answer 3

Here is a proof that mimics the proof of the maximum modulus principle.

Let $P$ be the polynomial and $f(x) = |P(x)|$. We know that $f: \mathbb{C} \to \mathbb{R}^+$ so it has a minimum. Assume for contradiction that this minimum is at $z$ where $P(z)\ne 0$

Recall the taylor series expansion $P(z+\epsilon) = P(z) + P'(z)\epsilon + P''(z)\epsilon^2 / 2 + \cdots + P^{(m)}(z) \epsilon^m / m! + \cdots$

Let $t$ be the smallest positive integer such that $P^{(t)}(z) \ne 0$. Then picking $|\epsilon| < 999^{-1000t}$ and $\frac{\epsilon^t P^{(t)}(z)}{P(z)}\in \mathbb{R}^-$ yields $$|P(z+\epsilon)| \le P(z) (1-|\frac{\epsilon^t P^{(t)}(z)}{P(z)t!}|) + \sum\limits_{j=t+1}^{\deg P} |\epsilon^{j} P^{j}(z) / j!|$$

By triangular inequality.

Picking $|\epsilon|$ sufficiently small gives $|P(z+\epsilon)| < |P(z)|$, contradicting $|P(z)|$'s minimality