# Homotopy cauchy

-If $D$ is convex (part of $\mathbb{C}$) and if we have two paths $\gamma_1$ and $\gamma_2$ in $D$ with $\gamma_1(a)=\gamma_2(a)$, and $\gamma_1(b)=\gamma_2(b)$. Proof that $\gamma_1$ and $\gamma_2$ are homotopic in $D$ as paths with constant endpoints. Intuitively it's clear, but how can I give an explicit homotopy? And if $D$ (part of $\mathbb{C}$, complex numbers) is open and $f: D \to \mathbb{C}$ holomorphic, how can I proof that $\int \limits_{\gamma_1} f(z)\:dz$ along path $\gamma_1$ is equal to $\int \limits_{\gamma_2} f(z)\:dz$ along path $\gamma_2$? (given that $\gamma_1$ and $gamma_2$ are homotopic in $D$ as paths with constant endpoints). I think that it has to do with the Cauchy theorem homotopic version.

## 1 Answer

• To construct the homotopy, note that for every $t$ the line segment $\ell_t$ joining $\gamma_1(t)$ to $\gamma_2(t)$ lies entirely in $D$. So the formula $H_s(t) = (1-s)\gamma_1(t) + s\gamma_2(t)$ is a homotopy of paths in $D$.

• Depending on what you mean by the "homotopic version of Cauchy's theorem", your question about integrals follows trivially from your first question. Read the statement of the theorem carefully, and if you're still stuck then write it out here.