So we seek the coefficient $a_{-1}$ of the Laurent series for $f(z) = \frac{z^2 + 1}{(z - i)^2}dz$, where Laurent series is denoted by

$$f(z) = \sum_{n = -\infty}^\infty a_n(z - z_0)^n$$

when the series is given about the point $z_0 = i$.

We know the coefficients of the Laurent series can be given by the integral

$$a_n = \frac{1}{2\pi i} \int_C \frac{f(z)}{(z - z_0)^{n+1}}dz =\frac{1}{2\pi i} \int_C \frac{z^2 + 1}{(z-i)^{n+3}}dz$$

where $C$ is a counterclockwise closed curve about $z_0 = i$. Here, the integrand doesn't have any other poles/singularities, so this curve need no other restrictions other than having nonzero radius.

Since we seek $a_{-1}$, we plug in $n = -1$:

$$a_{-1} = \frac{1}{2\pi i} \int_C \frac{z^2 + 1}{(z-i)^{2}}dz$$

Given the form of the integral matches the below,

$$\frac{1}{2\pi i} \int_C \frac{g(z)}{(z - z_0)^{m+1}}dz=\frac{1}{m!}g^{(m)}(z_0)$$

when $g(z)=z^2+1$, $z_0 = i$, and $m=1>0$, we can use this to calculate our integral, yielding

$$a_{-1} = \frac{1}{2\pi i} \int_C \frac{z^2 + 1}{(z-i)^{2}}dz= \frac{1}{1!}g'(i)=g'(i)$$

Since $g'(z)=2z$, then,

$$a_{-1} = 2i$$

I mostly just wanted to double-check this solution since I feel pretty weak when it comes to Laurent series.


2 Answers 2


Observe: $f(z) = \frac{z^2+1}{(z-i)^2} = \frac{(z+i)(z-i)}{(z-i)^2} = \frac{z+i}{z-i} = \frac{(z-i) +i +i}{z-i} = \frac{2i}{z-i} + 1.$ What is the coefficient for $a_{-1}$ given this observation?

  • $\begingroup$ I believe that, following from the definition of Laurent series, we would have $2i$ as well per this derivation, since $$\frac{2i}{z-i}=2i(z-i)^{-1}$$ with the power of $(z-i)$ hinting that this refers to the $a_{-1}$ term. And thus, similarly, if I wanted to find $a_0$, $a_0 = 1$. In which case I'd be correct, right? Just want to be sure. $\endgroup$ Nov 14, 2018 at 5:05
  • 1
    $\begingroup$ @EeveeTrainer indeed! $\endgroup$
    – user328442
    Nov 14, 2018 at 5:06

Just wanted to give some context to the already accepted answer:

In practice, when trying to compute the Laurent series for a complex function, we (almost) never use the definitions provided by the Laurent theorem to calculate the coefficients. Rather, we fool around with the Taylor series of known portions of the function until we get one sum with possibly negative powers of $z$, as in the accepted answer. This typically involves tricks like multiplying by 1, adding 0, and things like that. This in turn gives information about the integrals, according to the definitions $$ a_n = \frac{1}{2\pi i}\oint_{+\gamma}\frac{f(w)}{(w-z_0)^{n+1}}dw,\ \ b_n = \frac{1}{2\pi i}\oint_{+\gamma}f(w)(w-z_0)^{n-1}dw $$ where $\gamma$ is any curve defined in the annulus on which you are computing the Laurent series. This is well defined by the homotopy principle.

So in this case, we didn't need any Taylor series expansions, the expression happened to be nice enough to simplify to a Laurent series already.

  • $\begingroup$ Yeah I get that, it's still a mode of thought I'm trying to get into myself though since my professor waffles on the idea. Sometimes he seems to want the formal calculations of the coefficients, other times he wants us to do it the easy way, etc. I figured he specifically wanted the formal calculation in this case given a comment earlier in the evening (mostly just trying to contextualize this question/comment), though the alternate method was also helpful and illustrative as well. $\endgroup$ Nov 14, 2018 at 5:33

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