Prove that any closed path is $\mathbb C$-contractible. Prove that any closed path is $\mathbb C$-contractible.
I basically think that any closed path is homotopic to a circle, and any circle is $\mathbb C$-contractible. I am not sure whether there is a more detailed way to prove this. Any help?
Thanks~
 A: A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka  Exer 4.24a

I'll prove the case for simple paths $\gamma$ using Jordan Curve Theorem and expanding on your idea:


By Jordan Curve Theorem, $int(\gamma)$ is open, i.e. $\forall g \in int(\gamma), \exists r_g > 0 : D[g,r_g] \subseteq int(\gamma)$.
$\therefore, \gamma \sim_{\mathbb C}C[g,r_g]$. As you point out, $C[g,r_g]\sim_{\mathbb C}0$. I don't think it's explicit, but I think it's like in Figure 4.2
A: The book is an open textbook that is available at https://matthbeck.github.io/complex.html
On p.60, the term $G$-homotopic is defined for a region $G \subset \mathbb{C}$: two paths $f,g:[0,1] \to G$ are $G$-homotopic if they are homotopic in $G$, meaning there is a map $H:[0,1] \times [0,1] \to G$ such that $H(s,0) = f(s)$ and $H(s,1) = g(s)$.
A standard way to approach this problem when $G$ is convex (or star-like) region, e.g. the the entire complex plane $\mathbb{C})$, is to define a map $r:G \times [0,1] \to G$ such that $r(z,0) = z$ and $r(z,1) = z_0$, where $z_0$ is a chosen basepoint in $G$.  The so-called straight line homotopy works: $r(z,t) = (1-t)z + tz_0$.
Then you want to use the map $r$ to define a homotopy between a given path $f$ and the constant path at $z_0$.  Do the same for the second path $g$ and the constant path at $z_0$.  Finally, glue these homotopies (or appeal to the fact that $G$-homotopy defines an equivalence relation on paths if this has already been shown).
