Principal part of function at a pole

Question

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: $\dfrac{ze^z}{(z-1)(z+1)}=\sum_{k=0}^\infty \dfrac{z^kz}{k!(z+1)(z-1)}=\sum_{k=0}^\infty \dfrac{((z-1)+1)^k((z-1)+1)}{k!((z-1)+2)(z-1)} \\ =\sum_{k=0}^\infty \dfrac{\lbrack\sum_{n=0}^k\binom{k}{n}(z-1)^n\rbrack((z-1)+1)}{k!((z-1)+2)(z-1)}=\sum_{k=0}^\infty \dfrac{\lbrack\sum_{n=0}^k\binom{k}{n}(z-1)^{n-1}\rbrack((z-1)+1)}{k!((z-1)+2)} \\ =\sum_{k=0}^\infty \dfrac{\lbrack\sum_{n=0}^k\binom{k}{n}(z-1)^{n-1}\rbrack((z-1)+2)-\lbrack\sum_{n=0}^k\binom{k}{n}(z-1)^{n-1}\rbrack}{k!((z-1)+2)}= \\ \sum_{k=0}^\infty\dfrac{\sum_{n=0}^k\binom{k}{n}(z-1)^{n-1}}{k!}- \dfrac{\sum_{n=0}^k\binom{k}{n}(z-1)^{n-1}}{k!((z-1)+2)}$.

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)?

Answer 1

You are acting as if there should be a simple expression for that series. I doubt it. Consider the function $f\colon\mathbb{C}\setminus\{-1\}\longrightarrow\mathbb C$ 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.

Answer 2

You can rewrite your function as

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

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

$$\begin{align}e\frac{e^w(w+1)}{w(w+2)}&=\frac{e}w\left(1-\frac1{w+2}\right)\sum_{k=0}^\infty\frac{w^k}{k!}\\ &=\frac{e}w\left(1-\frac12\sum_{k=0}^\infty\left(-\frac{w}2\right)^k\right)\left(\sum_{k=0}^\infty\frac{w^k}{k!}\right)\\ &=e\sum_{k=0}^\infty\left(\sum_{j=0}^k\left(-\frac12\right)^{j+1}\frac{(-1)^{\delta_{j,0}}}{(k-j)!}\right)w^{k-1}\end{align}$$

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 $z=1$.

Answer 3

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: $$ \lim_{z\to 1} f(z)(z-1) = \lim_{z\to 1} \frac{e^z z}{z+1}=\frac{e}{2} $$