Finding the residue of $\frac{e^{zt}}{z^2(e^{\pi z}-1)}$ using the laurent series

Question

$$f(z)=\frac{e^{zt}}{z^2(e^{\pi z}-1)}\tag{1}$$

I am doing it as follows, but my textbook is convinced that i am wrong:

$ \color{lime}{1.\space Residue\space at\space z\space =0}$

let $\pi z := q$

$e^{q}-1 = q+\frac{q^2}{2}+\frac{q^3}{6}+..=q(1+\frac{q}{2}+\frac{q^2}{6}+..)\tag{a}$

$e^{zt} =1+ zt +\frac{(zt)^2}{2}+\frac{(zt)^3}{6}+.. \tag{b}$

using $(a)$ and $(b)$ we express $f(z)$ as follows:

$$f(z) = \frac{1}{qz^2}\{1+ z+\frac{(zt)^2}{2}+\frac{(zt)^3}{6}+..\}\{1+\frac{q}{2}+\frac{q^2}{6}+..\}^{-1}$$

re substituting $q$ $f(z)$ becomes:

$$f(z) = \frac{1}{\pi z^3}\{1+ zt +\frac{(zt)^2}{2}+\frac{(zt)^3}{6}+..\}\{1-(\frac{\pi z}{2}+\frac{(\pi z)^2}{6}+..)+(\frac{\pi z}{2}+\frac{(\pi z)^2}{6}+..)^2+..\}$$

grouping coefficients of $z^2$ to find $c_1$:

$z^2\{\frac{\pi ^2}{12} -\frac{t\pi}{2}+\frac{t^2}{2}\}\frac{1}{z^3\pi}$

$\therefore$ $\operatorname{Res(f(z);0)}=\frac{\pi}{12}-\frac{t}{2}+\frac{t^2}{2\pi}$

i did however manage to find the $ \color{lime}{\space Residue\space at\space z\space = 2ki}$

->can someone please confirm that "hopefully" my book is wrong - Thanks.

Answer 1

Since: $$ e^{zt} = 1+zt+\frac{t^2}{2}z^2+O(z^3),$$ $$\frac{z}{e^{\pi z}-1}=\frac{1}{\pi}-\frac{1}{2}z+\frac{\pi}{12}z^2+O(z^3) $$ we have: $$\begin{eqnarray*}\operatorname{Res}\left(\frac{e^{zt}}{z^2(e^{\pi z}-1)},z=0\right)&=&[z^2]\left(1+zt+\frac{t^2}{2}z^2\right)\cdot\left(\frac{1}{\pi}-\frac{1}{2}z+\frac{\pi}{12}z^2\right)\\&=&\color{red}{\frac{\pi}{12}-\frac{t}{2}+\frac{t^2}{2\pi}}\end{eqnarray*}$$ as you stated.

Answer 2

Using Maple I am obtaining that the residue at $z=0$ is

$${\frac {{t}^{2}}{2\pi }}+\frac{\pi}{12} -\frac{t}{2}$$

Then your result is correct.

For the residue at $z = 2ik$ I am obtaining

$$-{\frac {{{\rm e}^{2\,ikt}}}{4{k}^{2}\pi }}$$

Do you agree?