Cauchy's Integral Formula Examples 1

# Cauchy's Integral Formula Examples 1

Recall from the Cauchy's Integral Formula page that if $A \subseteq \mathbb{C}$ is open, $f : A \to \mathbb{C}$ is analytic on $A$, and $\gamma$ is a simple, closed, piecewise smooth positively oriented curve contained in $A$ then for all $z_0$ in the inside of $\gamma$ we have that the value of $f$ at $z_0$ is:

(1)
\begin{align} \quad f(z_0) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{z - z_0} \: dz \end{align}

We will now look at some example problems involving applying Cauchy's integral formula.

## Example 1

Evaluate the integrals $\displaystyle{\int_{\gamma} \frac{\sin z}{z} \: dz}$ and $\displaystyle{\int_{\gamma} \frac{\cos z}{z} \: dz}$ where $\gamma$ is the positively oriented circle centered at $0$ with radius $1$.

Note that $\displaystyle{\sin z}$ and $\displaystyle{\cos z}$ are both analytic on all of $\mathbb{C}$ and $\gamma$ is a simple, closed, piecewise smooth, positively oriented curve contained in $\mathbb{C}$, and so for $z_0 = 0$ in the inside of $\gamma$ we have that:

(2)
\begin{align} \quad \sin (0) = \frac{1}{2\pi i} \int_{\gamma} \frac{\sin z}{z} \: dz \quad \Leftrightarrow \quad \int_{\gamma} \frac{\sin z}{z} \: dz = 0 \end{align}
(3)
\begin{align} \quad \cos (0) = \frac{1}{2\pi i} \int_{\gamma} \frac{\cos z}{z} \: dz \quad \Leftrightarrow \quad \int_{\gamma} \frac{\cos z}{z} \: dz = 2\pi i \end{align}

## Example 2

Evaluate the integral $\displaystyle{\int_{\gamma} \frac{z^2 - 1}{z^2 + 1} \: dz}$ where $\gamma$ is the positively oriented circle centered at $0$ with radius $1$.

We must first use some algebra in order to transform this problem to allow us to use Cauchy's integral formula. We have that:

(4)
\begin{align} \quad z^2 + 1 = (z + i)(z - i) \end{align}

So for some $A, B \in \mathbb{C}$ we have that:

(5)
\begin{align} \quad \frac{1}{z^2 + 1} = \frac{A}{z + i} + \frac{B}{z - i} \quad \Leftrightarrow \quad \frac{A(z - i)}{z^2 + 1} + \frac{B(z + i)}{z^2 + 1} \end{align}

Therefore $1 = A(z - i) + B(z + i)$. Setting $z = i$ gives us that $1 = 2iB$ so $\displaystyle{B = \frac{1}{2i} = -\frac{i}{2}}$. Setting $z = -i$ gives us that $1 = -2iA$ so $\displaystyle{A = -\frac{1}{2i} = \frac{i}{2}}$. Therefore:

(6)
\begin{align} \quad \frac{1}{z^2 + 1} = \frac{1}{2} \left ( \frac{i}{z + i} - \frac{i}{z - i} \right ) \end{align}

Then:

(7)
\begin{align} \quad \frac{z^2 - 1}{z^2 + 1} = \frac{i}{2} \left ( \frac{z^2 - 1}{z + i} - \frac{z^2 - 1}{z - i} \right ) \end{align}

Hence:

(8)
\begin{align} \quad \int_{\gamma} \frac{z^2 - 1}{z^2 + 1} \: dz = \frac{i}{2} \left ( \int_{\gamma} \frac{z^2 - 1}{z + i} \: dz - \int_{\gamma} \frac{z^2 - 1}{z - i} \: dz \right ) \end{align}

Note that $f(z) = z^2 - 1$ is analytic on all of $\mathbb{C}$ and contains the simple, closed, piecewise smooth, positively oriented curve in $\mathbb{C}$. So by Cauchy's integral formula we have that:

(9)
\begin{align} \quad \int_{\gamma} \frac{z^2 - 1}{z^2 + 1} \: dz &= \frac{i}{2} \left ( 2\pi i f(-i) - 2\pi i f(i) \right ) \\ &= -\pi \left ( [(-i)^2 - 1] - [(i)^2 - 1] \right ) \\ &= -\pi \left ( (-1 - 1) - (-1 - 1) \right ) \\ &= -\pi \left ( (-2) - (-2) \right ) \\ &= 0 \end{align}