Understand this Residue

Question

To determine that residue, we recall (Sec. 64) the Maclaurin series representation $$ e^z=\sum_{n=0}^\infty\frac{z^n}{n!}\quad(|z|\lt\infty) $$ and use it to write $$ \frac{e^z-1}{z^5}=\frac1{z^5}\sum_{n=0}^\infty\frac{z^n}{n!}=\sum_{n=0}^\infty\frac{z^{n-5}}{n!}\quad(0\lt|z|\lt\infty). $$ The coefficient of $1/z$ in this last series occurs when $n-5=-1$, or when $n=4$. Hence $$ \operatorname*{Res}_{z=0}\frac{e^z-1}{z^5}=\frac1{4!}=\frac1{24}; $$ and so $$ \int_C\frac{e^x-1}{z^4}\,\mathrm{d}z=2\pi i\left(\frac1{24}\right)=\frac{\pi i}{12}. $$

Original Image

So I am reading this section on residues but I don't seem to quite understand why the Maclaurin series was evaluated for $\frac{e^z-1}{z^5}$ instead of $\frac{e^z-1}{z^4}$.

Remark: This is taken from "Complex Variables and Applications" - James Ward Brown and Ruel V. Churchill.

Answer 1

Clearly it is a typo. With the same method you can see that $$ \mbox{Res}_{z=0} \left( \frac{e^z-1}{z^4}\right)=\frac{1}{6} $$ so $$ \int_C \frac{e^z-1}{z^4} dz= 2\pi i \cdot \frac{1}{6} $$

Answer 2

It is a typo. If we change the last equation to $$ \int_C\frac{e^z-1}{z^\color{#C00}{5}}\,\mathrm{d}z=2\pi i\left(\frac1{24}\right)=\frac{\pi i}{12} $$ everything else is good.