Confusion to a solution to finding the Principal Part of a Laurent Series

Question

Hello I am trying to find the Principal part at the pole $z = 1$ of the function $\frac{z}{(z^2-1)^2}$. Clearly, the function has a pole at $z=1,-1$, and by a theorem, $z=1$ is an order $2$ pole. Now the principal part of a pole is defined in my book as

The sum of the negative powers, $$ P(z)=\sum_{k=-N}^{-1} a_{k}\left(z-z_{0}\right)^{k}=\frac{a_{-N}}{\left(z-z_{0}\right)^{N}}+\cdots+\frac{a_{-1}}{z-z_{0}}, $$ $\text { is called the principal part of } f(z) \text { at the pole } z_{0} \text {. }$, where $N$ is the order of the pole. so in my case,

$(1)$ $P(z) = \sum_{k=-2}^{-1} a_{k}\left(z-1\right)^{k}=\frac{a_{-2}}{(z-1)^2} + \frac{a_{-1}}{(z-1)}$ ?. How would I find the $a_k$ in this case? I went back to the definition of a Laurent Series and saw the definition $a_n$ is defined as $(2)$ $a_{n}=\frac{1}{2 \pi i} \oint_{\left|z-z_{0}\right|=r} \frac{f(z)}{\left(z-z_{0}\right)^{n+1}} d z, \quad-\infty<n<\infty$.

But this confuses me because the answer in the book is

$\text { Double poles at } \pm 1 \text {, principal parts } \mp(1 / 4)(z \pm 1)^{2} \text {. }$

so by looking at the answer, apparently $a_{-1} = 0$ since the answer does not have the $\frac{a_{-1}}{(z-1)}$ term in $(1)$. But if I used $(2)$ to find $a_{-1}$, it becomes $a_{-1}=\frac{1}{2 \pi i} \oint_{\left|z-1\right|=r} \frac{\frac{z}{(z^2-1)^2}}{\left(z-1\right)^{0}} d z$, which is not $0$ because its not analytic at the domain and cannot apply cauchy theorem.

Am I overthinking this problem? how can I find the $a_k$ in this problem?

Answer 1

I suppose you're thinking abit too much into it: The simplest way to find a Laurent expansion is by not calculating it at all.

Here's what I mean: Near $z=1$, you can view your function $f(z) = \frac{z}{(z^2-1)^2}$ as the product of two very different functions: $$f(z) = \frac{1}{(z-1)^2}\cdot\frac{z}{(z+1)^2}$$

The left factor is already written in its Laurent series, while the right factor is holomorphic around $z=1$, meaning it can be expressed as a regular power series (i.e. with only non-negative exponents): $$\frac{z}{(z+1)^2} = \sum\limits_{k=0}^\infty a_k(z-1)^k = a_0 + a_1(z-1)+a_2(z-1)^2...$$

So when multiplying both factors to get $f(z)$, we see immediatly that all terms $a_k(z-1)^k$ with $k\geq 2$ have non-negative exponents, i.e. don't belong to the principal part:

$$f(z) = \frac{1}{(z-1)^2}\cdot(a_0 + a_1(z-1)+a_2(z-1)^2+...) = a_0(z-1)^{-2}+a_1(z-1)^{-1}+a_2+...$$

Basically the exponents of the original series are shifted by $-2$. So to get the desired coefficients for your Laurent expansion of $f$ at $z=1$, you only have to calculate $a_0 = \left. \frac{z}{(z+1)^2}\right|_{z=1} = \frac{1}{4}$ and $a_1 = \left. \left(\frac{z}{(z^2+1)^2}\right)'\right|_{z=1} = 0$.

That $a_1$ vanishes is also evident from the formula you mentioned in your post: $$ \frac{1}{2\pi i}\int\limits_{\partial B_r(1)} \frac{f(z)}{(z-1)^0} d z = \frac{1}{2\pi i}\int\limits_{\partial B_r(1)} \frac{z}{(z-1)^2(z+1)^2} d z = 0,$$ since the integrand has no singularity of order $1$ inside the chosen curve of integration.

Answer 2

We can also make a Laurent series expansion at $z=1$ up to a point where we can see the principal part easily.

We obtain \begin{align*} \color{blue}{\frac{z}{\left(z^2-1\right)^2}}&=\frac{z}{(z-1)^2}\,\frac{1}{\left(2+(z-1)\right)^2}\\ &=\frac{z}{4(z-1)^2}\,\frac{1}{\left(1+\frac{z-1}{2}\right)^2}\\ &=\frac{(z-1)+1}{4(z-1)^2}\,\sum_{n=0}^\infty \binom{-2}{n}\left(\frac{z-1}{2}\right)^n\tag{1}\\ &\,\,\color{blue}{=\sum_{n=0}^{\infty}(-1)^n\frac{n+1}{2^{n+2}}(z-1)^{n-1}+\sum_{n=0}^{\infty}(-1)^n\frac{n+1}{2^{n+2}}(z-1)^{n-2}}\tag{2}\\ \end{align*}

Comment:

• In (1) we apply the binomial series expansion.

• In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

From (2) we can already get the principal part by taking the term with $n=0$ from the left-hand series and the terms with $n=0$ and $n=1$ from the right-hand series.

We obtain \begin{align*} &\frac{1}{4}(z-1)^{-1}+\frac{1}{4}(z-1)^{-2}-\frac{1}{4}(z-1)^{-1}\\ &\qquad\color{blue}{=\frac{1}{4}(z-1)^{-2}} \end{align*} in accordance with the solution to the problem.