Integrals: Complex Functions Along P.S. Curves Examples 1

Integrals of Complex Functions Along Piecewise Smooth Curves Examples 1

Recall from the Integrals of Complex Functions Along Piecewise Smooth Curves page that if $h : [a, b] \to \mathbb{C}$ is a single real-variable, complex-valued function where $u, v : [a, b] \to \mathbb{C}$ and such that $h(t) = u(t) + iv(t)$ then the integral of $h$ over $[a, b]$ is defined as:

(1)
\begin{align} \quad \int_{a}^{b} h(t) \: dt = \int_a^b u(t) \: dt + i \int_a^b v(t) \: dt \end{align}

Furthermore, if $A \subseteq \mathbb{C}$ is open, $f : A \to \mathbb{C}$, and $\gamma : [a, b] \to \mathbb{C}$ is a piecewise smooth curve contained in $A$, (i.e., there exists a partition $a = a_0 < a_1 < ... < a_n = b$ for which $\gamma$ exists on $(a_k, a_{k+1})$ and continuous on $[a_k, a_{k+1}]$ for all $k \in \{ 0, 1, ..., n - 1 \}$), then the integral of $f$ along $\gamma$ is defined as:

(2)
\begin{align} \quad \int_{\gamma} f(z) \: dz = \int_{a}^{b} f(\gamma(t)) \cdot \gamma'(t) \: dt = \sum_{k=0}^{n-1} \int_{a_k}^{a_{k+1}} f(\gamma(t)) \cdot \gamma'(t) \: dt \end{align}

We will now look at some examples of computing integrals of complex functions.

Example 1

Evaluate the integral $\displaystyle{\int_{\gamma} \mathrm{Re} (z) \: dz}$ where $\gamma$ is the line segment with initial point at the origin and with terminal point at $2 - i$.

The curve $\gamma$ be can be parameterized for $t \in [0, 1]$ by:

(3)
\begin{align} \quad \gamma (t) = (2 - i)t \end{align}

The derivative of $\gamma$ is:

(4)
\begin{align} \quad \gamma'(t) = 2 - i \end{align}

Therefore the integral of $\mathrm{Re} (z)$ along $\gamma$ is:

(5)
\begin{align} \quad \int_{\gamma} \mathrm{Re} (z) \: dz &= \int_0^1 \mathrm{Re} (\gamma (t)) \cdot \gamma'(t) \: dt \\ &= \int_0^1 (2t) \cdot (2 - i) \: dt \\ &= (4 - 2i) \int_0^1 t \: dt \\ &= (4 - 2i) \left [ \frac{t^2}{2} \right ]_{t=0}^{t=1} \\ &= \frac{4 - 2i}{2} \\ &= 2 - i \end{align}

Example 2

Evaluate the integral $\displaystyle{\int_{\gamma} \mathrm{Im} (z) \: dz}$ where $\gamma$ is the line segment with initial point at $1 + i$ and with terminal point at $-1 - i$.

The curve $\gamma$ can be parameterized for $t \in [0, 1]$ by:

(6)
\begin{align} \quad \gamma(t) &= t(-1 - i) + (1 - t)(1 + i) \\ &= (1 - 2t)(1 + i) \end{align}

The derivative of $\gamma$ is:

(7)
\begin{align} \quad \gamma'(t) &= -2(1 + i) \\ &= -2 - 2i \end{align}

Therefore the derivative of $\mathrm{Im} (z)$ along $\gamma$ is:

(8)
\begin{align} \quad \int_{\gamma} \mathrm{Im} (z) \: dz &= \int_0^1 \mathrm{Im} (\gamma(t)) \cdot \gamma'(t) \: dt \\ &= \int_0^1 \mathrm{Im} [(1 - 2t)(1 + i)] \cdot [-2 - 2i] \: dt \\ &= [-2 - 2i] \int_0^1 (1 - 2t) \: dt \\ &= [-2 - 2i] \left [ t - t^2 \right ]_{t=0}^{1} \\ &= 0 \end{align}