An analytical proof for the punctured plane is not simply connected? I am trying to make a "challenge problem" for my (undergraduate) real analysis students. Currently, the students knows about connectedness, compactness in $\mathbb R^n$, functional limits and continuous functions from $\mathbb R^m$ to $\mathbb R^n$. They are also familiar with all the standard real analysis results on $\mathbb R$. 
The goal for this problem is to show that the punctured plane is not simply connected. I set up the problem by defining the terms "path", "loop", "path homotopy", and "null-homotopic". But I could not come up with a proof accessible to real analysis students showing that the path parameterized by $\alpha(t)
 = (\cos(2 \pi t),\sin(2 \pi t))$ is NOT null-homotopic. Can someone help?
Things I definitely want to avoid: fundamental groups, Brouwer fixed point theorem, residue theorem. 
Things I wish to avoid: There is a proof using Green's theorem, which I guess has the same flavor as the residue theorem in complex analysis. I think this is something students are able to understand. But since we have not talked about vector calculus in this course, it would be better if the proof I write down as solution does not involve Green's theorem.
 A: You could invoke the winding number. For a curve $t\mapsto {\bf z}(t)$ in the right half plane you have the running polar angle
$$\phi(t)=\arctan{y(t)\over x(t)}$$
and therefore $$\phi'(t)={x(t)y'(t)-y(t)x'(t)\over x^2(t)+y^2(t)}\ .$$
Your students will believe (or easily check) that this formula remains valid in the full punctured plane. It follows that the total polar angle increment $\Delta\phi$ along a closed curve $$\gamma:\quad t\mapsto {\bf z}(t)\ne{\bf 0}\qquad(a\leq t\leq b)$$
is given by
$$\Delta\phi=\int_a^b \phi'(t)\>dt=\int_a^b {x(t)y'(t)-y(t)x'(t)\over x^2(t)+y^2(t)}\>dt\ .$$
(If you do not want to assume $\gamma$ smooth you can define $\Delta\phi$ using  finite partitions of the interval $[a,b]$.) This $\Delta\phi$ has to be an integer multiple of $2\pi$, hence the winding number
$$N(\gamma,{\bf 0}):={1\over2\pi}\int_a^b \phi'(t)\>dt=\int_a^b {x(t)y'(t)-y(t)x'(t)\over x^2(t)+y^2(t)}\>dt\tag{1}$$
is an integer. 
As the RHS of $(1)$ is a continuous function of not shown "homotopy time" it follows that $N(\gamma,{\bf 0})$ is invariant under a homotopy. Now your curve $\alpha$ has $N(\alpha,{\bf 0})=1$, while a small circle $\gamma$ around the point ${\bf e}:=(1,0)$ obviously has $N(\gamma,{\bf e})=0$.
