I have been reading through "(adsbygoogle = window.adsbygoogle || []).push({}); Complex Analysis for Mathematics & Engineering"byJ. 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 theoppositecurve ##-\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 $$-\mathcal{C}:\;z_{2}(t)=x(t)+iy(t)\,,\;\;\; b\leq t\leq a$$

Furthermore, how does one show that if two contours ##\mathcal{C}_{1}## and ##\mathcal{C}_{2}## are placed end to end, such that the terminal point of ##\mathcal{C}_{1}## coincides with the initial point ##\mathcal{C}_{2}##, then the integral over the contour ##\mathcal{C}=\mathcal{C}_{1}+\mathcal{C}_{2}## has the property $$\int_{\mathcal{C}}f(z)dz=\int_{\mathcal{C}_{1}+\mathcal{C}_{2}}f(z)dz= \int_{\mathcal{C}_{1}}f(z)dz+\int_{\mathcal{C}_{2}}f(z)dz$$ Would one use the following parametrisations: $$\mathcal{C}_{1}:\;z(t)=x(t)+iy(t)\,,\;\;\; a\leq t\leq b\\ \mathcal{C}_{2}:\;z(t)=x(t)+iy(t)\,,\;\;\; b\leq t\leq c$$ Then, $$\int_{\mathcal{C}_{1}}f(z)dz+\int_{\mathcal{C}_{2}}f(z)dz=\int_{a}^{b}f(z(t))z'(t)dt+\int_{b}^{c}f(z(t))z'(t)dt=\int_{a}^{c}f(z(t))z'(t)dt=\int_{\mathcal{C}_{1}+\mathcal{C}_{2}}f(z)dz=\int_{\mathcal{C}}f(z)dz$$ where $$\mathcal{C}:\;z(t)=x(t)+iy(t)\,,\;\;\; a\leq t\leq c$$

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# I Contour integration - reversing orientation

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