Consider a contour $\mathcal{C}$ $$\mathcal{C}:[a,b]\longrightarrow\mathbb{C}\,,\;\;\;$$ Then the opposite curve $-\mathcal{C}$ traces out the same set of points but in the reverse order defined as $$-\mathcal{C}(t)=\mathcal{C}(a+b-t)\;\;\;a\leq t\leq b.$$ Suppose the function
f and g are continuous complex functions of a complex variable in a domain $D$ and $\mathcal{C}$ is lying entire in D, then $$\int_{C} f(z) dz= -\int_{-C} f(z) dz$$. Can any one explain stepwise how?
My attempt is $\int_{-C} f(z) dz=\int_{a}^b f((-C)(t)){(-C)'(t)} dt= \int_{a}^b f(C(a+b+t)){C'(a+b+t)}dt$.
I cannot proceed past this to get the desired result.
Best Answer
Let us write $D$ for the reversed path, in order to avoid confusion with the minus signs. Then $$ D(t) = C(a+b-t) \\ D'(t) = -C'(a+b-t) $$ and therefore $$ \int_D f(z) \,dz = \int_a^b f(D(t))D'(t) \, dt = \int_a^b f(C(a+b-t) (-C'(a+b-t)) \, dt \, . $$ Now substitute $s=a+b-t$, $ds = -dt$: $$ \cdots = \int_b^a f(C(t))C'(t) \, dt = -\int_a^b f(C(t))C'(t) \, dt = -\int_{-C} f(z) \,dz \, . $$