<s>[INST] Determine the residuals of $f(z)=\frac{\log(1+z)}{\cos z-1}$ Determine the residuals of $f$
$f$ is given by
$$
f(z)=\frac{\log(1+z)}{\cos z-1}, \quad z \in \mathbb{C}-(-\infty, -1]
$$
where $\log z$ is the principal logarithm in the cutplane $\mathbb{C}_\pi:=\mathbb{C}-\{z \in \mathbb{R} \mid z \leq 0\}$.
Determining the poles:
Since
$$
\cos z-1=0 \Leftrightarrow \cos z=1 \Leftrightarrow z=2\pi k,
\quad k \in \mathbb{Z}
$$
the set of poles of $f$ is given by
$$
P=\{2\pi k \mid k \in \mathbb{N} \cup \{0\}\}
$$
For $z=0$ we can write $f$ as
$$
f(z)=\frac{\log(1+z)}{\cos z-1}
=
\frac{z-\frac{z^2}{2}+\frac{z^3}{3}-\cdots}{-\frac{z^2}{2!}+\frac{z^4}{4!}- \cdots}
=
\frac{z \left(1-\frac{z}{2}+\frac{z^2}{3}- \cdots \right)}
{z^2 \left(\frac{1}{2!}+\frac{z^2}{4!}- \cdots \right)}
=
\frac{\left(1-\frac{z}{2}+\frac{z^2}{3}- \cdots \right)}
{z \left(\frac{1}{2!}+\frac{z^2}{4!}- \cdots \right)}
$$
and therefore,
$$
(z-0)f(z)=
\frac{\left(1-\frac{z}{2}+\frac{z^2}{3}- \cdots \right)}
{\left(\frac{1}{2!}+\frac{z^2}{4!}- \cdots \right)}
$$
The limit of this as $z \rightarrow 0$ is $2$, so the residue in $z=0$ is 2.
But how do I determine the other residuals?
Here's one of my attemps
Since cosine is periodic we can write $f$ as
$$
f(z)=\frac{\log(1+z)}{\cos z-1}=\frac{\log(1+z)}{\cos(z-2\pi k)-1}
=
\frac{\left(z-\frac{z^2}{2}+\frac{z^3}{3}- \cdots \right)}
{(z-2\pi k)^2\left(\frac{1}{2!}+\frac{(z-2\pi k)^2}{4!}- \cdots \right)}
$$
and therefore,
$$
(z-2\pi k)^2f(z)=
\frac{\left(z-\frac{z^2}{2}+\frac{z^3}{3}- \cdots \right)}
{\left(\frac{1}{2!}+\frac{(z-2\pi k)^2}{4!}- \cdots \right)}
$$
However, taking the derivative of this is not particularly nice...
Any suggestions?
 [/INST] I'll do it when $k=1$. The general case is similar.
You have\begin{align}\frac{\log(1+z)}{\cos(z)-1}&=\frac{\log(1+2\pi+z-2\pi)}{\cos(z-2\pi)-1}\\&=\frac{\log(1+2\pi)+\frac{z-2\pi}{1+2\pi}-\frac{(z-2\pi)^2}{2(1+2\pi)^2}+\cdots}{-\frac{(z-2\pi)^2}{2!}+\frac{(z-2\pi)^4}{4!}-\cdots}\\&=\frac1{(z-2\pi)^2}\times\frac{\log(1+2\pi)+\frac{z-2\pi}{1+2\pi}-\frac{(z-2\pi)^2}{2(1+2\pi)^2}+\cdots}{-\frac12+\frac{(z-2\pi)^2}{4!}-\cdots}\end{align}Clearly, this quotient can be written as$$\frac{a_{-2}}{(z-2\pi)^2}+\frac{a_{-1}}{z-2\pi}+a_0+a_1(z-2\pi)+\cdots$$and the residue is $a_{-1}$. But\begin{multline}\log(1+2\pi)+\frac{z-2\pi}{1+2\pi}-\frac{(z-2\pi)^2}{2(1+2\pi)^2}+\cdots=\\=\left(-\frac12+\frac{(z-2\pi)^2}{4!}-\cdots\right)\times\left(a_{-2}+a_{-1}(z-2\pi)+a_0(z-2\pi)^2+\cdots\right)=\\=-\frac{a_{-2}}2-\frac{a_{-1}}2(z-2\pi)+\cdots\end{multline}and therefore $-\frac{a_{-1}}2=\frac1{1+2\pi}$. So$$\operatorname{res}_{z=2\pi}\frac{\log(1+z)}{\cos(z)-1}=a_{-1}=-\frac2{1+2\pi}.$$
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