Cauchy's Integral Theorem Examples 1

Recall from the Cauchy's Integral Theorem page the following two results:

The Cauchy-Goursat Integral Theorem for Open Disks:

• If $f$ is analytic on an open disk $D(z_0, r)$ then for any closed, piecewise smooth curve $\gamma$ in $D(z_0, r)$ we have that:

(1)

\begin{align} \quad \int_{\gamma} f(z) \: dz = 0 \end{align}

• Furthermore, there exists a function $F$ that is analytic on $D(z_0, r)$ such that $F'(z) = f(z)$.

The Cauchy Integral Theorem:

• If $A \subseteq \mathbb{C}$ is open, $f : A \to \mathbb{C}$ is analytic on $A$, and if $\gamma$ is a closed, piecewise smooth curve in $A$ homotopic to a point in $A$ then:

(2)

\begin{align} \quad \int_{\gamma} f(z) \: dz =0 \end{align}

We will now use these theorems to evaluate some seemingly difficult integrals of complex functions. That said, it should be noted that these examples are somewhat contrived. In practice, knowing when (and if) either of the Cauchy's integral theorems can be applied is a matter of checking whether the conditions of the theorems are satisfied.

Example 1

Evaluate the integral $\displaystyle{\int_{\gamma} \frac{e^z}{z} \: dz}$ where $\gamma$ is given parametrically for $t \in [0, 2\pi)$ by $\gamma(t) = e^{it} + 3$.

Note that the function $\displaystyle{f(z) = \frac{e^z}{z}}$ is analytic on $\mathbb{C} \setminus \{ 0 \}$. The curve $\gamma$ is the circle of of radius $1$ shifted $3$ units to the right. This circle is homotopic to any point in $D(3, 1)$ which is contained in $\mathbb{C} \setminus \{ 0 \}$. So by Cauchy's integral theorem we have that:

(3)

\begin{align} \quad \int_{\gamma} \frac{e^z}{z} \: dz = 0 \end{align}

Example 2

Consider the function $\displaystyle{f(z) = \left\{\begin{matrix} z^2 & \mathrm{if} \: \mid z \mid \leq 3 \\ \mid z \mid & \mathrm{if} \: \mid z \mid > 3 \end{matrix}\right.}$ and let $\gamma$ be the unit square. Evaluate $\displaystyle{\int_{\gamma} f(z) \: dz}$.

Note that $f$ is analytic on $D(0, 3)$ but $f$ is not analytic on $\mathbb{C} \setminus D(0, 3)$ (we have already proved that $\mid z \mid$ is not analytic anywhere). So since $f$ is analytic on the open disk $D(0, 3)$, for any closed, piecewise smooth curve $\gamma$ in $D(0, 3)$ we have by the Cauchy-Goursat integral theorem that $\displaystyle{\int_{\gamma} f(z) \: dz = 0}$. In particular, the unit square, $\gamma$ is contained in $D(0, 3)$. Thus:

(4)

\begin{align} \quad \displaystyle{\int_{\gamma} f(z) \: dz} = 0 \end{align}