It is easy to see that a circle and a semi-circle (as in the image), of the same radius, are homotopic. I was trying to exhibit a homotopy, but it seems that I'll have to use analytic geometry, solve second degree equations, etc, in the way. I am lazy and I do not want to do that, if it can be avoided.
Is there an intelligent way of doing this?
Context: I want to prove that if $R > 1$, then the index of $\gamma: [0, 2\pi] \to \Bbb C$ given by $\gamma(t) = Re^{it}$ if $0 \leq t \leq \pi$, and $\gamma(t) = \frac{2R}{\pi}(t-\pi)-R$ if $\pi \leq t \leq 2\pi$ with respect to $i$ is $1$. I can use that a homotopy preserves the index, and if a point $\alpha$ has absolute value less than $R$, then ${\rm Ind}(\sigma, \alpha) = 1$, if $\sigma(t) = Re^{it}$, $0\leq t \leq 2\pi$.
You can use a straight line homotopy, $$ (t,(x,y))\mapsto \left\{ \begin{array}{cc} (x,t\sqrt{1-x^2})&y\geq0\\ (x,y)&y<0\\ \end{array} \right. $$ between the two maps $S^1\to\mathbb{R}^2$.