I'm a high school student, curious about proofs of the Fundamental Theorem of Algebra. Specifically, I've been thinking about one of the topological proofs of the theorem, given in Courant's book, "What is Mathematics".
Here are my questions:
- But the order $φ(t)$ depends continuously on $t$, since $f(z)$ is a continuous function of $z$. Hence we shall have a contradiction, for the function $φ(t)$ can assume only integral values and therefore cannot pass continuously from the value $0$ to the value $n$.
I don't get this part. What is it trying to say? I know that $f(z)$ is a continuous function, and that $φ(t)$ changes 'smoothly' with $t$ (in my mind I have a picture of a closed curve shrinking 'smoothly' when the circle that $z$ traces also shrinks in the same way, hence, at every value of $t$, there exists a unique order. However, I also know that the order must have integral values, because there can't be $3.14$ turns.) But is this correct? I think I'm getting these concepts in a muddle. I also can't see how this contradiction serves to show that our initial assumption about $f(z)$ not having a root is absurd.
- Since the expression on the left is the distance between the two points $z^n$ and $f(z)$, while the last expression on the right is the distance of the point $z^n$ from the origin, we see that the straight line segment joining the two points $f(z)$ and $z^n$ cannot pass through the origin so long as z is on the circle of radius $t$ about the origin.
What does this really imply? In my mind I think that whenever z traces out a circle with a radius $t$ $\neq$ $0$, the rest of the terms in $f(z)$ will never be $0$. But how does this help with proving that when t is large, $f(z)$ behaves like $z^n$?
- This being so, we may continuously deform the curve traced out by $f(z)$ into the curve traced out by $z^n$ without ever passing through the origin, simply by pushing each point $f(z)$ along the segment joining it to $z^n$. Since the order of the origin will vary continuously and can assume only integral values during this deformation, it must be the same for both curves. Since the order for $z^n$ is $n$, the order for $f(z)$ must also be $n$.
What does "deforming $f(z)$ into $z^n$ mean, and why do we want to do this? How can the order vary continuously if it can only assume integral values? How does this show that the orders of $z^n$ and $f(z)$ are the same?
Thanks, this is my first time posting here and I'm not well-versed with a lot of the terms in Mathematics (I'll be finishing high school in six months) so forgive my sloppy way of explaining things.
This is actually a very sloppy simplification of an important and complicated proof of the fundamental theorem of algebra, which secretly proves nontriviality of $\pi_1(S^1)$ behind the curtains (which involves some nontrivial bit of algebraic topology). I'll try to give a more explicit description of what's really going on in the proof by answering your questions :
Once we know that when $r$ is large, the curve $\Gamma_f$ traced by $f(z)$ ($z$ traversing the circle of radius $r$) looks like $z^n$ (which can be visualized as a loop winding $n$ times around the origin) and when $r = 0$, $\Gamma_f$ is some point other than the origin, the contradiction comes naturally -- as $\Gamma_f$ varies continuously with $r$, these two "states" at large $r$ and $r = 0$ must've been obtained continuously, and then $\Gamma_f$ must have crossed the origin at least once, which implies $f(z) = 0$ for some $z$. To prove this rigorously, one develops the notion of "order", commonly known as $\mathsf{winding \, number}$. As order of the origin for any closed curve is fixed under perturbations of the curve, it must also be fixed under the continuous process $\Gamma_f$ went through. However, this doesn't seem to happen : order of $\Gamma_f$ for $r= 0$ is $0$, while on the other hand it's $n$ for large $r$. That's the general idea of the proof.
It's not enough to know that $f(z)$ behaves like $z^n$ for large $r$, really. What we want to show is that order of $\Gamma_f$ for large $r$ is $n$. To do this, note as before that order is fixed under small perturbations of the curve. We must, thus, construct a small perturbation of the curve $\Gamma_f$ away from the origin ('cause we are counting the order of our curve with respect to that!). A little algebra (as in the article) shows that $|f(z) - z^n| < |z|^n$ which means the distance between $f(z)$ and $z^n$ is small enough to slide $f(z)$ to $z^n$ without passing through the origin.
You've put your finger on the matter, in truth. I have never defined what a continuous deformation is anywhere. I've been patching this up with vague notion of "small perturbations", but now you've got me. The correct formalization of this idea is really fundamentals of algebraic topology, but I'll try to describe it in a restricted codomain : a $\mathsf{loop}$ in the complex plane $\Bbb C$ is a continuous function $\gamma : [0, 1] \to \Bbb C$ such that $\gamma(0) = \gamma(1)$. You can visualize this as taking the open interval, mapping it continuously in the complex plane, and then gluing the two endpoints. So the terminology makes sense. Now, if you have two loops $\gamma_1, \gamma_2 : [0, 1] \to \Bbb C$, then a continuous deformation between them is a class of continuous loops $\{f_t : [0, 1] \to \Bbb C\}$ where $t \in [0, 1]$ such that $f_0 = \gamma_1$ and $f_1 = \gamma_2$ (the two "states"). This completely formalizes the idea of small perturbations (I recommend drawing a few pictures to really grasp the idea). In fact, this is more commonly known as a $\mathsf{homotopy}$ between the two loops. You can actually construct an explicit homotopy between the curve traced by $f(z)$ and $z^n$ respectively (and indeed, a homotopy without crossing the origin), and it's a fact that winding numbers are homotopy-invariant, i.e., homotopic loops have the same winding numbers. So the whole point of all this is that it really proves that the curve $\Gamma_f$ traced by $f(z)$ has order $n$.
I hope this clears up your doubts. I have left a lot of unexplained terminology in the answer, and it's not expected that you'll understand concepts such as homotopy, but it's worth a shot, I guess.