Unit circle traversed once in the punctured plane not homotopic to the same circle traversed twice. On page 185 of Saff, Fundamentals of Complex Analysis, the author states that in the punctured plane ($\mathbb{C}- \{0\}$), the unit circle traversed once in the positive direction is NOT continuously deformable to the unit circle traversed twice in the positive direction. 
I can't understand why the deformation $z(s,t):[0,1]^{2} \rightarrow \mathbb{C} - \{0\}$ given by $$z(s,t) = e^{2\pi t(1+s)i}$$ doesn't do the trick. For $s=1$ don't we get twice the speed?  
 A: I think you misunderstand the statement.  Any path in $\mathbb C \smallsetminus \{0\}$ can be homotoped to any other path if the endpoints may be moved, since $\mathbb C\smallsetminus\{0\}$ is connected.  For the statement about circles to be interesting, it is important to interpret it in terms of circles, or, equivalently, in terms of continuously deforming paths without moving either endpoint.
So the question is to prove that the function $S^1 = \mathbb R / \mathbb Z \to \mathbb C$ given by sending $\theta \mapsto \exp(2\pi i\theta )$ is not homotopic to $\theta \mapsto \exp(4\pi i \theta)$.  There are many proofs of this fact.  I will give one that I think is quite informative, but without a lot more work applies only to smooth deformations between smooth paths.  (By approximating other paths, this proof can be used to conclude the statement about continuous deformations of continuous paths, but I will not do so.)  The reason I like this proof is that it uses the complex numbers.
The basic fact to be used is that the contour integral of a meromorphic 1-form $\alpha = f(z)\mathrm d z$ along a curve that avoids all poles of $f$ is invariant under endpoint-fixing deformations of the curve; if the endpoints move, then the integral changes by something depending only on the endpoints.  In particular, for any curve $\gamma : S^1 \to \mathbb C$ avoiding the poles of $f$, $\oint_\gamma f(z)\mathrm dz$ is invariant under small changes to $\gamma$.  It thus suffices to find a 1-form (holomoprhic on $\mathbb C\smallsetminus \{0\}$, with a pole at $0$) that distinguishes the two curves in question.
The function $f(z)\mathrm dz = z^{-1}\mathrm d z$ does the trick.  Its integral around the pole is just $2\pi i$, whereas if you go twice around you get $4\pi i$.
