Laurent series for $f (z)=\frac {\sin (2 \pi z)}{z (z^2 + 1)}$ How Can find the Laurent series for this function valid for $0 <|z-i|<2$ $$f (z)=\frac {\sin (2 \pi z)}{z (z^2 +1)}$$
Let $g (z) = \sin (\pi z)$
$$\sin (\pi z ) = \sin( 2 \pi  (z - i)) \cos (2 \pi i) + \cos (2 \pi  (z-i)) \sin (2 \pi i )$$
And Let $h (z)= \frac {1}{z^2 + 1}$
$$\frac {1}{z (z^2 + 1)}= \frac {1}{i (1 -(-(z-i))}[\frac {1/2i}{z-i} +\frac {-1/2i}{2i (1-(-\frac {z-i}{2i}))}]$$
So it's easy to find expansion for $g (z)$ and $h (z)$ and then multiply the two expansions 
We notice that $ f $ has simple pole at $z = i$ So, we can get the principal part easily Or using this 
$$2 \pi i a_1 = \int_{|z-i|=1} f (z) dz$$  
Is there a trick to find the Laurent series quickly ?
This question was in my exam .I Calculated the principal part , but I didn't have enough time to calculate the exact form for the analytic part .
Thank you 
 A: First we obtain
$$\dfrac{1}{z(z+i)}=\dfrac{i}{z+i}-\dfrac{i}{z}=\dfrac{1}{2\left(1+\dfrac{z-i}{2i}\right)}-\dfrac{1}{\left(1+\dfrac{z-i}{i}\right)}={\sum_{k=0}^\infty(-1)^k\left\{\dfrac{1}{2}\cdot\dfrac{1}{(2i)^k}-\dfrac{1}{i^k}\right\}\cdot(z-i)^k}.$$
The Taylor series converges for $\vert z-i\vert\lt1$. Next we obtain
$$\sin(2\pi z)=\sin\{2\pi(z-i)+2\pi i\}=
{\sin\{2\pi(z-i)\}\cos(2\pi i)+\cos\{2\pi(z-i)\}\sin(2\pi i)}=
{\cos(2\pi i)\sum_{k=0}^\infty(-1)^k\dfrac{\{2\pi (z-i)\}^{2k+1}}{(2k+1)!}+\sin(2\pi i)\sum_{k=0}^\infty(-1)^k\dfrac{\{2\pi (z-i)\}^{2k}}{(2k)!}}=
{\cos(2\pi i)\sum_{k=0}^\infty\sin\dfrac{k\pi}{2}\cdot\dfrac{\{2\pi (z-i)\}^{k}}{k!}+\sin(2\pi i)\sum_{k=0}^\infty\cos\dfrac{k\pi}{2}\cdot\dfrac{\{2\pi (z-i)\}^{k}}{k!}}=
{\sum_{k=0}^\infty\sin\left\{\dfrac{k\pi}{2}+2\pi i\right\}\dfrac{\{2\pi (z-i)\}^k}{k!}}.$$
This Taylor series converges on $\mathbb C$. The product gives
$$\tag{1}\dfrac{\sin(2\pi z)}{z(z+i)}=\sum_{j=0}^\infty a_j(z-i)^j$$
with
$$a_j:=\sum_{k=0}^j (-1)^k\left\{\dfrac{1}{2}\cdot\dfrac{1}{(2i)^k}-\dfrac{1}{i^k}\right\}\cdot\left\{\sin\left[\dfrac{(j-k)\pi}{2}+2\pi i\right]\cdot\dfrac{(2\pi)^{j-k}}{(j-k)!}\right\}.$$
The product converges for the smallest radius of convergence which is $\vert z-i\vert\lt1$. But the function
$$\tag{2}\dfrac{\sin(2\pi z)}{z(z+i)}$$
is holomorphic for $\vert z-i\vert\lt2$. Therefore the product converges even for $\vert z-i\vert\lt2$. Now we only have to multiply $(1)$ with $(z-i)^{-1}$.
Supplement:
My understanding of the comments is that I multiply two absolutely convergent series, Taylor series in this case, and the product does not converge or it does not converge on the given radius of convergence. Unfortunately I only read German books about analysis. Therefore I will only give German references. But you are free to edit English references.
First we obtain: Theorem $32.6$, Lehrbuch der Analysis, Teil $1$, Harro Heuser [$H1$], $2009$: If the series $\sum_{k=0}^\infty a_k$ and $\sum_{k=0}^\infty b_k$ are absolutely convergent, their Cauchy product
$$\sum_{k=0}^\infty (a_0b_k+a_1b_{k-1}+\dots+a_kb_0)=\left(\sum_{k=0}^\infty a_k\right)\cdot\left(\sum_{k=0}^\infty b_k\right)$$
is absolutely convergent. Hence within the radius of convergence of two Taylor series we are allowed to take their Cauchy product and that is what I did in $(1)$. Explicitely this is stated for Taylor series in theorem $63.3$, [$H1$]. The latter theorem states that the radius of convergence is the minimum of the the radii of each of the taylor series.
Now complex analysis comes into play. I have established that the series $(1)$ converges for $\vert z-i\vert\lt1$. It is a Taylor series of the function $(2)$. By the identity theorem of Taylor series there exists only one Taylor series for a function, theorem $64.5$, [$H1$] (if there exists one at all). Now complex analysis states: Theorem $187.2$, Lehrbuch der Analysis, Teil $2$, Harro Heuser [$H2$], $2008$: A holomorphic function can be developed to a series $\sum a_k(z-z_0)^k$ around a point $z_0$ in the open neighbourhood $G$ where it is defined. It will converge at least in the largest open disc with center $z_0$ that still lies in $G$. Because this disc can be established to $\vert z-i\vert\lt2$ the series $(1)$ must even converge on this disc!
In my steps of deduction I first stated that the Taylor series $(1)$ holds for $\vert z-i\vert\lt1$. Without this step I may not extend the radius of convergence if I have not proven that it is the taylor series at all. Please keep this in mind. All hinges on the identity theorem of Taylor series.
