Complex Line Integrals 1: Complex Line Integrals

Let $f(z): \mathcal{R}\longrightarrow\mathbb{C}$ be a piecewise continuous function defined in a domain $\mathcal{R}$ and let $C$ be the differentiable curve $\gamma(t): [a,b]\longrightarrow\mathcal{R}$. Here we assume that $\gamma(t)$ is differentiable on an open interval containing $[a,b]$. The complex line integral of $f(z)$ along the curve $C$, $\int_Cf(z)dz$ is defined to be
$$
\int_Cf(z)dz=\int_a^bf(\gamma(t))\gamma'(t)dt
$$

The complex line integral of $f(z)$ along the curve $C$ satisfies the following properties.

Theorem.
\begin{align*}
\int_Cz_0f(z)dz&=z_0\int_Cf(z)dz \tag{1}\\
\int_C[f(z)+g(z)]dz&=\int_Cf(z)dz+\int_Cg(z)dz \tag{2}
\end{align*}
These two properties (1) and (2) imply that $\int_Cf(z)dz$ is linear.

Let $-C$ denote the same curve as $C$ with opposite orientation. Let $C$ be $\gamma(t): [a,b]\longrightarrow\mathcal{R}$. Then $-C$ is defined by $\gamma(\tau)$ where $\tau=a+b-t$.
\begin{align*}
\int_{-C}f(z)dz&=\int_b^a f(\gamma(\tau))\gamma'(\tau)d\tau\\
&=-\int_a^bf(\gamma(t))\gamma'(t)dt\\
&=-\int_Cf(z)dz
\end{align*}
Hence, we have
$$
\int_{-C}f(z)dz=-\int_Cf(z)dz
$$

Suppose that $C$ is a path that consists of a path $C_1$:  $\gamma(t)$, $a\leq t\leq c$ followed by a path $C_2$: $\gamma(t)$, $c\leq t\leq b$. Then
\begin{align*}
\int_{C_1}f(z)dz+\int_{C_2}f(z)dz&=\int_a^cf(\gamma(t))\gamma'(t)dt+\int_c^bf(\gamma(t))\gamma'(t)dt\\
&=\int_a^bf(\gamma(t))\gamma'(t)dt\\
&=\int_Cf(z)dz
\end{align*}
Hence, we have
$$
\int_Cf(z)dz=\int_{C_1}f(z)dz+\int_{C_2}f(z)dz
$$

Example. Evaluate the integral $\int_C\bar z dz$ where $C$ is the right half of the circle $|z|=2$.

Solution. Let $C$: $\gamma(\theta)=2e^{i\theta}$, $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$. Then
\begin{align*}
\int_C\bar z dz&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\overline{2e^{i\theta}}(2e^{i\theta})'d\theta\\
&=4i\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\theta=4\pi i
\end{align*}

Remark. The result in the preceding example can be used to evaluate $\int_C\frac{1}{z}dz$ where $C$ is the same path in the example.
\begin{align*}
\int_C\frac{1}{z}dz&=\int_C\frac{\bar z}{z\bar z}dz\\
&=\int_C\frac{\bar z}{|z|^2}dz\\
&=\frac{1}{4}\int_C\bar z dz\ (|z|=2)\\
&=\pi i
\end{align*}

Example. Let $C_1$ denote the path $OAB$ and $C_2$ denote the path $OB$ in figure 1. 

Figure 1. Path dependence

Evaluate the integrals $\int_{C_1}f(z)dz$ and $\int_{C_2}f(z)dz$, where $f(z)=y-x-i3x^2$ ($z=x+iy$).

Solution. The path $OAB$ can be represented by
$$
\gamma(t)=\left\{\begin{array}{ccc}
it & \mbox{if} & 0\leq t\leq 1\\
(t-1)+i & \mbox{if} & 1\leq t\leq 2
\end{array}
\right.
$$
Thus,
\begin{align*}
\int_{C_1}f(z)dz&=\int_0^1f(\gamma(t))\gamma'(t)dt+\int_1^2f(\gamma(t))\gamma'(t)dt\\
&=i\int_0^1tdt+\int_1^2[2-t-i3(t-1)^2]dt\\
&=\frac{1-i}{2}
\end{align*}
Alternatively, the path $OA$ can be represented by $z=0+iy$, $0\leq y\leq 1$, while the path $AB$ can be represented by $z=x+i$, $0\leq x\leq 1$. Hence, we have
\begin{align*}
\int_{C_1}f(z)dz&=\int_{\overline{OA}}f(z)dz+\int_{\overline{AB}}f(z)dz\\
&=i\int_0^1ydy+\int_0^1(1-x-i3x^2)dx\\
&=\frac{1-i}{2}
\end{align*}
$C_2$ can be represented by $z=x+ix$, $0\leq x\leq 1$ and
\begin{align*}
\int_{C_2}f(z)dz&=\int_0^1-i3x^2(1+i)dx\\
&=3(1-i)\int_0^1x^2dx\\
&=1-i.
\end{align*}
As seen in this example, while the two paths $C_1$ and $C_2$ have the same initial and terminal points, the integrals $\int_{C_1}f(z)dz$ and $\int_{C_2}f(z)dz$ have different values. This means that the line integrals are in general depend on the paths.