complex integral

[logistic_guy]

Solve $\displaystyle \int_{C}\frac{e^z - 1}{z^4} \ dz$, where $\displaystyle C$ is the positively oriented unit circle $\displaystyle |z| = 1$ as shown in the figure.

 

[logistic_guy]

$\displaystyle \int_{C}\frac{e^z - 1}{z^4} \ dz = 2\pi i \ \underset{z = 0}{\text{Res}}\left(\frac{e^z - 1}{z^4}\right)$

If we can just find the residue at $\displaystyle z = 0$ for the integrand, we are done!

In the next post, we will try to figure out a way to find the residue with the help of the Maclaurin series.

 

[logistic_guy]

The Maclaurin expansion for $\displaystyle e^z$ is:

$\displaystyle e^z = 1 + z + \frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24} + \frac{z^5}{120} + \cdots$

 

[logistic_guy]

The Maclaurin expansion for $\displaystyle e^z - 1$ is:

$\displaystyle e^z - 1 = z + \frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24} + \frac{z^5}{120} + \cdots$

 

[logistic_guy]

When we divide by $\displaystyle z^4$, we get:

$\displaystyle \frac{e^z - 1}{z^4} = \frac{1}{z^3} + \frac{1}{2z^2} + \frac{1}{6z} + \frac{1}{24} + \frac{z}{120} + \cdots$

 

[logistic_guy]

When we divide by $\displaystyle z^4$, we get:

$\displaystyle \frac{e^z - 1}{z^4} = \frac{1}{z^3} + \frac{1}{2z^2} + \frac{1}{6z} + \frac{1}{24} + \frac{z}{120} + \cdots$

This Laurent series tells us that the residue at $\displaystyle z = 0$ is:

$\displaystyle \frac{1}{6}$

We are almost there

 

[logistic_guy]

Here are the final touches.

$\displaystyle \int_{C}\frac{e^z - 1}{z^4} \ dz = 2\pi i \ \underset{z = 0}{\text{Res}}\left(\frac{e^z - 1}{z^4}\right) = 2\pi i \ \frac{1}{6} = \frac{\pi i}{3}$