Cfrgtkky

I have a function $dfrac{e^zz}{z^2-1}$. It has isolated singularities $z=pm 1$.
To find the principal part at $z=1$, I am trying to find a Laurent series expansion around $z=1$. I have the following mess:
$dfrac{ze^z}{(z-1)(z+1)}=sum_{k=0}^infty dfrac{z^kz}{k!(z+1)(z-1)}=sum_{k=0}^infty dfrac{((z-1)+1)^k((z-1)+1)}{k!((z-1)+2)(z-1)}
\
=sum_{k=0}^infty dfrac{lbracksum_{n=0}^kbinom{k}{n}(z-1)^nrbrack((z-1)+1)}{k!((z-1)+2)(z-1)}=sum_{k=0}^infty dfrac{lbracksum_{n=0}^kbinom{k}{n}(z-1)^{n-1}rbrack((z-1)+1)}{k!((z-1)+2)}
\
=sum_{k=0}^infty dfrac{lbracksum_{n=0}^kbinom{k}{n}(z-1)^{n-1}rbrack((z-1)+2)-lbracksum_{n=0}^kbinom{k}{n}(z-1)^{n-1}rbrack}{k!((z-1)+2)}=
\
sum_{k=0}^inftydfrac{sum_{n=0}^kbinom{k}{n}(z-1)^{n-1}}{k!}- dfrac{sum_{n=0}^kbinom{k}{n}(z-1)^{n-1}}{k!((z-1)+2)}$.

How can I get a Laurent series just with powers of $(z-1)$ so that I can find the principal part at $z=1$ (that is, the part of the series with negative powers)?

You are acting as if there should be a simple expression for that series. I doubt it. Consider the function $fcolonmathbb{C}setminus{-1}longrightarrowmathbb C$ defined by $f(z)=frac{ze^z}{z+1}$. The first terms of its Taylor series are$$frac e2+frac{3e}4(z-1)+frac{3e}8(z-1)^2+frac{7e}{48}(z-1)^3+cdots,$$from which you can deduce that the Laurent series that you're after is$$frac e{2(z-1)}+frac{3e}4+frac{3e}8(z-1)+frac{7e}{48}(z-1)^2+cdots,$$but that's all.

You can rewrite your function as

$$frac{e^{w+1}(w+1)}{(w+1)^2-1}=efrac{e^w(w+1)}{w(w+2)}tag1$$

where we make the change $w+1=z$. Hence the Laurent series of your original function around $z=1$ is the Laurent series of the expression in $(1)$ around zero, that is

$$begin{align}efrac{e^w(w+1)}{w(w+2)}&=frac{e}wleft(1-frac1{w+2}right)sum_{k=0}^inftyfrac{w^k}{k!}\
&=frac{e}wleft(1-frac12sum_{k=0}^inftyleft(-frac{w}2right)^kright)left(sum_{k=0}^inftyfrac{w^k}{k!}right)\
&=esum_{k=0}^inftyleft(sum_{j=0}^kleft(-frac12right)^{j+1}frac{(-1)^{delta_{j,0}}}{(k-j)!}right)w^{k-1}end{align}$$

where $delta_{0,j}$ is the Kronecker delta. Then we only need to substitute $w=z-1$ in the last expression to have the form of the Laurent series of your original function around $z=1$.

The function $f(z)=frac{e^z z}{z^2-1}$ can be written as $frac{1}{z-1}$ times $frac{e^z z}{z+1}$, this last function being holomorphic in a neighbourhood of $z=1$. Therefore in $z=1$ the function $f$ has a pole of order $1$. This means that the Laurent series of the function $f$ around $z=1$ can be written as:
$$
f(z)=frac{a_{-1}}{z-1}+a_0+a_1(z-1)+a_2(z-1)+dots
$$
Now, if you are interested only in the part with negative powers, all you want to know is the value of $a_{-1}$, which can be evaluated via the limit:
$$
lim_{zto 1} f(z)(z-1) = lim_{zto 1} frac{e^z z}{z+1}=frac{e}{2}
$$