I have been reading through "Complex Analysis for Mathematics & Engineering" by J. Matthews and R.Howell, and I'm a bit confused about the way in which they have parametrised the opposite orientation of a contour $\mathcal{C}$.
Using their notation, consider a contour $\mathcal{C}$ with parametrisation $$\mathcal{C}:\;z_{1}(t)=x(t)+iy(t)\,,\;\;\; a\leq t\leq b$$ Then the opposite curve $-\mathcal{C}$ traces out the same set of points but in the reverse order, and it has the parametrisation $$-\mathcal{C}:\;z_{2}(t)=x(-t)+iy(-t)\,,\;\;\; -a\leq t\leq -b$$ They then go on to say that since $z_{2}(t)=z_{1}(-t)$, it is easy to see that $-\mathcal{C}$ is merely $\mathcal{C}$ traversed in the opposite sense.
However, I don't quite see why the contour $-\mathcal{C}$ is parametrised this way? Naively, I would've though it would be something like $$-\mathcal{C}:\;z_{2}(t)=x(t)+iy(t)\,,\;\;\; b\leq t\leq a$$
Best Answer
I think you have a typo and the reverse parameterization should read: $$ -\mathcal{C}:\;z_{2}(t)=x(-t)+iy(-t)\,,\;\;\; -b\leq t\leq -a $$
For $-b\leq t\leq -a$ we have $a\leq -t\leq b$ and $z_2(t) = x(-t) + iy(-t) = z_1(-t)$
And $z_2(-b) = z_1(b)\,, \; z_2(-a) = z_1(a)$ so $z_2$ traverses the same contour in the opposite direction from $z_1$
With regards to the naive intuition you mentioned, you can't have $b\leq t\leq a$ because you know that $a \lt b$ assuming they are different. Intuitively you want the parameterization to give the same values as $z_1(t)$ but when $t$ goes from $b$ to $a$, which is exactly what the book is doing.