I want to create a homotopy between the paths $$I \to S^{1} \quad \text{by} \quad t \mapsto e^{i \pi + 2 i \pi t} $$ and $$ I \to S^{1} \quad \text{by} \quad t \mapsto \begin{cases} e^{i \pi (1-4t)}, & t \in [0, \frac{1}{4}]\\ e^{i \pi (8t - 2)}, & t \in [\frac{1}{4}, \frac{1}{2}]\\ e^{i \pi (2t-1)},& t \in [\frac{1}{2}, 1] \end{cases}. $$
I'm having a bit of trouble getting all of the components to fit in, but visually, I think what one wants is to make the second path run through $[0, \frac{3}{8}]$ in less and less time until it becomes that it doesn't run through that part at all. Then runs through $[\frac{3}{8}, \frac{1}{2}]$ in eventually $[0, \frac{1}{2}]$, and the rest runs through in the same time. If that makes sense...
Could anyone help me along in understanding how to create this type of mapping?
Edit: Woops, I forgot the 2 in $i \pi + 2 i \pi t$ in the first path's expression. Thanks for pointing it out @Martin Brandenburg
Call your two functions $f(t)=e^{i\pi \alpha(t)}$ and $g(t)=e^{i\pi\beta(t)}$.
If you graph $\alpha,\beta$ in the plane, the first is a line segment, the second is three line segments. Notice the third line segment of $\beta$ differs from $\alpha$ by $2\pi$ $-$ thus, we can replace the third formula for $\beta(t)$ with $2t+1$ instead of $2t-1$, without changing the function $g(t)$. Now $\alpha,\beta$ match on $[\frac{1}{2},1]$.
It remains to homotope $\beta$ to $\alpha$ on $[0,\frac{1}{2}]$. Note the graph of $\beta$ is made up of two line segments, first from $(0,1)$ to $(\frac{1}{4},0)$ and then from $(\frac{1}{4},0)$ to $(\frac{1}{2},2)$; just "pull" the middle point up until the two line segments become the same, i.e. that of $\alpha$! Do you see how this can be done with formulas?