Now that we know what the fundamental group of the circle is . We can use this knowledge to prove the fundamental theorem of algebra!

The theorem states that if is a polynomial over with a degree greater than zero, then there exists such that .

This is actually kind of wierd. So far, I haven’t even **mentioned** polynomials at all! How can we prove this theorem with paths, and fundamental groups and all those stuff that I’ve discussed about so far…

It turns out that the property of having a root is in a way, a topological property. How exactly? We’ll see exactly how.

Before I begin with this theorem, I just want to mention that this theorem has a lot of proofs. One of them uses **galois theory** – which I am currently writing about. Another two proofs that I know come from **complex analysis**. I intend to present them all. So this post is one of **four** that discusses the fundamental theorem of algebra.

So I want to make some preperations before the proof that will make the proof smooth and staright-forward.

### The map

What this map does? Let’s see where it maps an arbitrary complex number . I’ll present in his **polar form**:

z=Re^{i\theta}

Thus:

z^n=(Re^{i\theta})^n=R^ne^{in\theta}

Ok, it’s pretty clear what’s happennig, it increases the **norm**: from to . And multiplies the argument by .

For example, if Then the map will map to .

Ok, here is some more interesting case: If we raise to the power of the number we’ll get

Re^{i2\pi}\mapsto R^ne^{i2\pi n}

For example, If the output will be . It’s not so clear how it’s a better example, but I think the animation will explain it better:

Applying this map causes this specific point to circle the origin 4 times. And in general, times.

My point is, that if we apply this map on the **loop**:

\varphi_1(t)=re^{2\pi it}

We will get:

\varphi_1(t)=re^{2\pi it}\mapsto {(re^{2\pi it})}^n=r^ne^{2\pi int}

So if a loop circles the origin once over a circle with radius , then after the map applied, the loop will now circle the origin times over a circle with radius .

### Homotopy in the complex plane

Suppose that such that:

|f(x)-g(x)|<|g(x)|

Then is homotopic to .

Why? we can easily define the homotopy:

H(x,t)=(1-t)\cdot g(x)+t\cdot f(x)

However, who said that for all ? If there are values such that this term is zero, then the homotopy is not even defined! Since is not in the range!

However, suppose that there are such values, and for those values:

(1-t)\cdot g(x)+t\cdot f(x)=0\\ \Downarrow \\ -\frac{(1-t)}{t}\cdot g(x)=f(x)

Thus:

|f(x)-g(x)|=|-\frac{(1-t)}{t}\cdot g(x)-g(x)|=|(\frac{t-1}{t}-1 )\cdot g(x)|

=|\frac{-1}{t} \cdot g(x)|=\frac{1}{t} \cdot |g(x)|\overset{0\leq t\leq1}{\geq}|g(x)|

And that’s a contradiction! (if , we’ll also get similar contradiction)

### Relation between homomorphism

Suppose that , . Thus:

f_*=0\iff g_*=0

This follows from the fact that if two paths are homotopic then there is an **isomorphism** (which I’ve defined here) such that:

g_*=F_\gamma\circ f_*

And from that relation, the statement is clearly true! (Make sure you understand why)

## The proof

Yes! it’s finally time for the proof!

Suppose that and:

p(z)=z^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0\ \ , \ \ n>0

I can assume that the leading coefficient is 1, cause if not, I can just divide the polynomial by it’s leading coefficient and that won’t affect the roots.

Aiming for contradiction, suppose that for every . Therefore, we can restrict the range of to . That is:

p:\mathbb{C}\to \mathbb{C}-\{0\}

Now, for every we denote:

S_r=\{z:|z|=r\}\ \ , \ \ D_r=\{z:|z|\leq r\}

Those are circles and disks. Consider the restriction:

p|_{S_r}:S_r\to\mathbb{C}-\{0\}

Now notice that this is a map from a circle that can be expanded to the map:

p|_{D_r}:D_r\to\mathbb{C}-\{0\}

which is defined all over the disk.

Recall that this property was one of 3 equivalent properties (right here). From the other two, we can conclude that is **null-homotopic**.

Great, now I’ll define another function:

g:\mathbb{C}\to\mathbb{C-\{0\}}\\g(z)=z^n

Notice that:

p(z)-g(z)=a_{n-1}z^{n-1}+\cdots+a_1z+a_0

Therefore:

\frac{|p(z)-g(z)|}{|g(z)|}=\frac{|a_{n-1}z^{n-1}+\cdots+a_1z+a_0|}{|z|^n}

We now use the triangle inequality to get:

\leq \frac{|a_{n-1}||z|^{n-1}+\cdots+|a_1||z|+|a_0|}{|z|^n}=\frac{|a_{n-1}|}{|z|}+\cdots+\frac{|a_1|}{|z|^{n-1}}+\frac{|a_0|}{|z|^n}\underset{|z|\to0}{\to}\infty

Thus, there exists large enough such that on :

\frac{|p|_{S_r}(z)-g|_{S_r}(z)|}{|g|_{S_r}(z)|}<\frac{|a_{n-1}|}{|z|}+\cdots+\frac{|a_1|}{|z|^{n-1}}+\frac{|a_0|}{|z|^n}<1 \\ \Downarrow \\ |p|_{S_r}(z)-g|_{S_r}(z)|<|g|_{S_r}(z)|

But we proved that this implies:

p|_{S_r}\sim g|_{S_r}

However, is null-homotopic, thus:

p|_{S_r}\sim K_b\Rightarrow {p|_{S_r}}_*={K_b}_*=0

Therefore:

{g|_{S_r}}_*=0

(This is just what I’ve mentioned here). On the other hand,

g|_{S_r}: S_r\to \mathbb{C}-\{0\} \\ \Downarrow \\ {g|_{S_r}}_*:\underbrace{\pi_1(S_r,r)}_{\mathbb{Z}}\to\underbrace{\pi_1(\mathbb{C}-\{0\},r^n)}_{\mathbb{Z}}

(Recall that and we’ve seen that ) Now, consider the loop :

{g|_{S_r}}_*([\varphi_1])={z^n|_{S_r}}_*([\varphi_1])=[z^n\circ\varphi_1]

We’ve already discussed what is. It is a loop that circles the origin times over a circle with radius . It’s equivalence class corresponds to and since , we got a **contradiction** to the map being trivial!

So we can’t restrict ‘s range to , hence, there is some such that .

## Summary

So the fact the polynomial’s degree is greater than zero, combined with the fact that the fundamental groups of the circle and are both led us to the contradiction!

That’s exactly the reason I think that proof is so great, we are not even using algebraic arguments. We don’t care about actually finding roots, factorize the polynomial – none of that! For some reason, the only thing that matter is the number of times a loop goes around the origin… Amazing!