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?
