I am Reading Complex Variables and Applications by Brown and Churchill. On page 230, the author defines the residue of the function $$f$$ at the isolated singular point $$z_0$$ as the coefficient of $$1/(z-z_0)$$ in the Laurent series expansion of $$f$$. The author then derives the following formula:

$$\int_C f(z)dz=2\pi i\underset{z=z_0}{\operatorname{Res}}f(z)$$

Simple enough. However, the author then gives the following example:

$$\int_C \frac{e^z-1}{z^4}dz=2\pi i\underset{z=0}{\operatorname{Res}}\left(\frac{e^z-1}{z^5}\right)$$

where $$C$$ is the positively oriented unit circle. Notice that the $$z^4$$ in the integral has become a $$z^5$$ in the residue. The author does not mention this and even calculates the residue correctly using $$z^5$$. Is this a mistake, or is there something going on here that I should be aware of? I am new to this stuff, so it is hard for me to tell what is correct and what is not.

• What is the residue the author gets? If it is $z^4$ then the residue should be $1/3!$. If it is $z^5$ then the residue should be $1/4!$. Commented Apr 25, 2021 at 1:56
• @JackozeeHakkiuz $1/4!$
– user772759
Commented Apr 25, 2021 at 1:57
• Then it should be $z^5$ everywhere. See the answer below. Commented Apr 25, 2021 at 2:01

We have for $$n\geq 1$$, \begin{align} \int_C\frac{e^z-1}{z^n}\,dz &= 2\pi i\cdot \text{Res}\left(\frac{e^z-1}{z^n}; 0\right)=\frac{2\pi i}{(n-1)!}. \end{align} So, one of the sides has a typo.