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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.

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  • $\begingroup$ 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 $1/4!$. $\endgroup$
    – Jackozee Hakkiuz
    Commented Apr 25, 2021 at 1:56
  • $\begingroup$ @JackozeeHakkiuz $1/4!$ $\endgroup$
    – user772759
    Commented Apr 25, 2021 at 1:57
  • $\begingroup$ Then it should be $z^5$ everywhere. See the answer below. $\endgroup$
    – Jackozee Hakkiuz
    Commented Apr 25, 2021 at 2:01

1 Answer 1

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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.

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