# Double checking a coefficient for a Laurent series, the series for $f(z) = \frac{z^2 + 1}{(z - i)^2}dz$

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.

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?
• 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. Nov 14, 2018 at 5:05
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.