Line integral.

A path of integration is a parameterized plane curve $C=\{ (x(t),y(t)), \; t \in [a,b] \}$ , where the functions $x(t)$ and $y(t)$ are continuous and have continuous first derivatives for $t \in [a,b]$ . We have:

$$\forall t \in [a,b], \; z(t)= x(t)+iy(t) dz = (x'(t)+iy'(t))\; dt$$

Let $f(z)=u(x,y)+iv(x,y)$ be a function continuous at every point of $C$ .

Definition 5.1.1       

$\int_C f(z) \; dz = \int_a^b [(u(x(t),y(t))+iv(x(t),y(t))](x'(t)+iy'(t))\; dt$ .

Example 5.1.2   Let $C$ be given by $x(t)=\cos t, y(t)= \sin t$ , where $t\in [0,\pi]$ , i.e. $C$ is the upper half unit-circle.

Figure 1:

Then $x'(t)= -\sin t$ and $y'(t)= \cos t$ . we have:

$$\int_C z \; dz = \int_0^{\pi} (\cos t + i \sin t ) ( -\sin t +i \cos t ) \; dt$$

$$\quad = \int_0^{\pi} ( -2 \sin t \cos t +i (\cos^2 t - \sin ^2 t) \; dt$$

$$\quad = \int_0^{\pi} ( - \sin 2t +i \cos 2t) \; dt$$

$$\quad = \left[ \frac 12 \cos 2t + \frac i2 \sin 2t \right]_0^{\pi} = 0.$$

Example 5.1.3   Let us compute the integral of the function $z \mapsto \overline{z}$ along two different paths.

First let $C$ be the same path as in previous example. If $z=\cos t + i \sin t$ , then $\overline{z}=\cos t - i \sin t$ . and we have:

$$\int_C \overline{z} \; dz = \int_0^{\pi} (\cos t - i \sin t ) ( -\sin t +i \cos t ) \; dt$$

$$\quad = \int_0^{\pi} i \; dt$$

$$\quad = \pi i .$$

Now we compute the integral of $z \mapsto \overline{z}$ along the segment from $1$ to $-1$ on the $x-$ axis; a parametrization of the segment is given by $z=-t$ , for $-1 \leq t \leq 1$ . Thus we have:

$$\int_C \overline{z} \; dz = \int_{-1}^{1} (-t)(-1) \; dt = \int_{-1}^{1} t \; dt = \left[ \frac 12 t^2 \right]_{-1}{1} =0.$$

Note that the two paths have the same origin and the same endpoint, but the integrals are different. We will understand this phenomenon later (v.i. Corollary 3.4).

Proposition 5.1.4       

1. The integral is independent of the choice of the parameterization.
2. $\int_C (f+g)(z) \; dz = \int_C f(z) \; dz + \int_C g(z) \; dz$ .
3. $\int_{C_1 \cup C_2} f(z) \; dz = \int_{C_1} f(z) \; dz + \int_{C_2} f(z) \; dz$ , when $C_1$ and $C_2$ are two paths such that the endpoint of $C_1$ and the origin of $C_2$ are identical.
4. If $C_1$ and $C_2$ are opposite paths, $\int_{C_1} f(z) \; dz - \int_{C_2} f(z) \; dz$ .