> [!NOTE] > Given a function $f: \Omega \subset \mathbb{C} \rightarrow \mathbb{C}$ along the path $\Gamma \subset \Omega \subset \mathbb{C}$ parametrised by $\gamma:[a, b] \rightarrow \mathbb{C}$ the integral of $f$ over $\Gamma$ is given by > > $\int_{\Gamma} f \mathrm{~d} z=\int_a^b f(\gamma(t)) \gamma^{\prime}(t) \mathrm{d} t=\int_a^b \boldsymbol{\operatorname { R e }}\left(f(\gamma(t)) \gamma^{\prime}(t)\right) \mathrm{d} t+\mathrm{i} \int_a^b \boldsymbol{\operatorname { I m }}\left(f(\gamma(t)) \gamma^{\prime}(t)\right) \mathrm{d} t$ Notice that we are not making any regularity assumptions on $f$, just that the integrals are well defined. Sometimes we will consider more than one parametrisation of a curve $\Gamma$, say $\gamma_1$ and $\gamma_2$ and will use the notation $\int_{\gamma_1} f$ and $\int_{\gamma_2} f$ in addition to $\int_{\Gamma}$. On many occasions we want to consider curves that are not $C^1$ but perhaps just piece-wise $C^1$. For example a square. In this case we can think of $\Gamma$ as a union of $n$ curves $\Gamma_j$, each one $C^1$, and parametrised in the right direction, so that connected in the right order they describe the entire curve $\Gamma$. We can define $ \int_{\Gamma} f \mathrm{~d} z:=\sum_{j=1}^n \int_{\Gamma_j} f \mathrm{~d} z $ It is straight forward from the definition that given a curve $\Gamma$, and two functions $f, g: \mathbb{C} \rightarrow \mathbb{C}$ and $\alpha, \beta \in \mathbb{C}$ we have $ \int_{\Gamma}(\alpha f(z)+\beta g(z)) \mathrm{d} z=\alpha \int_{\Gamma} f(z) \mathrm{d} z+\beta \int_{\Gamma} g(z) \mathrm{d} z $ If we allow for $\gamma^{\prime}(t)$ not to exists at finitely many points, this can be defined as a single integral, with clearly both formulations being equivalent. > [!Example] > Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be given by $f(z)=f(x+\mathrm{i} y)=x^4+\mathrm{i} y^4$ and the curve joining the origin in a straight line to the point $1+\mathrm{i}$, parametrized by $\gamma:[0,1] \rightarrow \mathbb{C}, \gamma(t)=(1+\mathrm{i}) t$. Notice that $\gamma^{\prime}(t)=1+\mathrm{i}$ and so we have $\int_{\Gamma} f=\int_0^1\left(t^4+\mathrm{i} t^4\right)(1+\mathrm{i}) \mathrm{d} t=\int_0^1 2 \mathrm{i}^4 \mathrm{~d} t=\frac{2}{5} \mathrm{i}$