"Let $[\alpha,\beta] (-\infty < \alpha\leq \beta)$ be a closed bounded interval in $\mathbb{R}$.
A $\mathbf{curve}\, \gamma$ with parameter interval $[\alpha,\beta]$ is a continuous function $\gamma: [\alpha,\beta]\rightarrow \mathbb{C}$. It has initial point $\gamma(\alpha)$ and final point $\gamma(\beta)$ and is closed if $\gamma(\alpha) = \gamma(\beta)$. It is simple if $\alpha \leq s \leq t \leq \beta$ implies tbat $\gamma(s) \neq \gamma(t)$ unless $s=t$, or in the case of a closed curve, $s = \alpha$ and $t = \beta$."
This is the definition of a $\textbf{curve}$ in my textbook "Introduction to Complex Analysis" by H.A. Priestley.
I am just getting more and more confused on all I though I knew! Here are the things I don't get:
- A curve is a continuous function. How can it be a function in the first place? There are examples in the book where I can take a line say $Re(z) = a$ in the Real axis, for some value $a$ and this will intersect the graph in more than one point!
It is parametrised. If it it parametrised, then how can it go back and forward, intersect and do all those "illegal" things? I thought that by parametrising it, I could use the values of $t$ from $\alpha$ to $\beta$ only once to sketch the graph! But if the curve crosses itself and goes ahead and behind, how can it be parametrised?
- The book then speaks about paths, and finally contours. And we want to integrate over contours, which are closed and formed by straight lines and arcs. We want to do so cause the integral along a closed contour is zero from what I could get. (not sure, it is quite vague) But then if $\mathbb{C}$ is a generalisation of $\mathbb{R}$ why doesn't this happen also in $\mathbb{R}$? I thought that only the integral of a conservative vector field along a closed path was equal to zero.