Recall that an arc is made up of the trace, the image of points, and the parametrisation, a way of driving along the curve.
Suppose C is a contour given by z(t) for t \in [a,b] with z_1 = z(a) and z_2 = z(b). Suppose f is piecewise continuous on C.
The contour integral is linear: \begin{aligned} \int_C (\alpha f)(z)\,dz &= \alpha \int_C f(z)\,dz \quad\text{ for }\alpha \in \mathbb C\\ \int_C(f+g)(z)\,dz &= \int_C f(z)\,dz + \int_Cg(z)\,dz \end{aligned}
C_1 + C_2 defines a contour only when the end of C_1 coincides with the start of C_2.
-C is defined as w(t) = z(-t) for -b \le t \le -a. Using the change of parameter formula, we can show that \int_{-C}f(z)\,dz = -\int_C f(z)\,dz. Hence, C_1 - C_2 = C_1 + (-C_2). This means the end of C_1 has to be the start of -C_2 (i.e. the end of C_2).
Example: Evaluate I = \int_C \bar z\,dz where C is given by z(\theta) = 2e^{i\theta} for -\pi/2 \le \theta \le \pi/2. This traces the right half of a circle with radius 2 counter-clockwise. We check that C is continuous on C (indeed, differentiable) and f is continuous on C. Note that z'(\theta) = 2ie^{i\theta}. Then, \begin{aligned} \implies I &= \int_{-\pi/2}^{\pi/2} f(z(\theta)) z'(\theta)\,d\theta \\ &= \int_{-\pi/2}^{\pi/2}\overline{(2e^{i\theta})}2ie^{i\theta}\,d\theta \\ &= 4i\int_{-\pi/2}^{\pi/2}{e^{-i\theta}}e^{i\theta}\,d\theta \\ &= 4i\int_{-\pi/2}^{\pi/2}\,d\theta \\ &= 4\pi i \end{aligned} On C, z \bar z = 4 which implies \bar z = 4/z. As a corollary, \int_C \frac{dz}z = \pi i. See §45 (8 Ed §41) for more examples.
Let D be a domain in \mathbb C (that is, an open connected subset of \mathbb C).
Definition. An antiderivative of f on D is F such that F'(z) = f(z) on D.
Theorem. The following three are equivalent:
Proof. (i) to (ii) follows from the fundamental theorem of calculus. For (ii) to (iii), take a closed contour C in D with z(a) = z(b) = z_1. Fix \gamma \in (a,b) such that z(\gamma) \ne z_1. Split C into two contours: C_1 with t \le \gamma and C_2 with t \ge \gamma. Then, C_1 + C_2 = C and \begin{aligned} \int_C f &= \int_{C_1 + C_2} f = \int_{C_1}f + \int_{C_2} f = \int_{C_1} - \int_{-C_2} f = 0, \end{aligned} because -C_2 and C_1 have the same start and end points so their integrals are equal by (ii). For (iii) to (ii) to (i), see B/C. \square
In particular, for C from z_1 \to z_2 in D, it holds that \int_C f(z)\,dz = F(b) - F(a), for any antiderivative F of f.
Keep in mind that we are doing integration, which is more of an art than a science. That is, it can be very difficult to get a (closed) for solution for even simple-looking integrands.
Example 2: I = \int_0^{1 + i} z^2\,dz. Here, f(z) = z^2 has an antiderivative, such as F(z) = z^3/3. By the FToC, I = F(1 + i) - F(0) = \frac{2}3(-1 + i). Example 3: I = \int_C dz / z^2, with C = 2e^{i\theta} and 0 \le \theta \le 2\pi. The integrand 1/z^2 has an antiderivative on \mathbb C_*, namely -1/z. Because C is a closed contour lying completely within \mathbb C_*, (iii) implies I = 0.
More generally, the same argument shows that \int_C z^n\,dz = 0 for all closed contours C and n \in \mathbb Z \setminus \{-1\}.