TITLE: Laurent series for $f (z)=\frac {\sin (2 \pi z)}{z (z^2 + 1)}$
QUESTION [8 upvotes]: 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

REPLY [3 votes]: Let us consider the function
$$f(t) = t^{-1/2}e^{2\pi it^{1/2}} = g_0(h(t))e^{2i\pi t^{1/2}},$$
where
$$g_0(u) = {1\over u},\quad h(t) = t^{1/2},\quad h'(t) = {1\over2}t^{-1/2} = {1\over2u}.$$
Let 
$$f^{(n)}(t) = g_n(h(t))e^{2\pi it}\tag1,$$
then
$$f^{(n)}(t) = \left(g_{n-1}(h(t))e^{2\pi it}\right)'h'(t) = \left(g'_{n-1}+2\pi ig_{n-1}\right)e^{2i\pi t}{1\over2u},$$
$$g_n = {1\over2u}g'_{n-1}+{i\pi\over u}g_{n-1}.\tag2$$
For example,
$$g_1(u) = {1\over2u}\left(-{1\over u^2}\right) + {i\pi\over u^2} = -{1\over2u^3} + {i\pi\over u^2},$$
$$g_2(u) = {1\over2u}\left(-{1\over2u^3} + {i\pi\over u^2}\right)' + {i\pi\over u}\left(-{1\over2u^3} + {i\pi\over u^2}\right)$$
$$ ={1\over2u}\left({3\over2u^4} - {2i\pi\over u^3}\right) + \left(-{i\pi\over2u^4} - {\pi^2\over u^3}\right) = \left({3\over4u^5} - {3\pi i\over2u^4} - {\pi^2\over u^3}\right),$$
$$g_3(u) = {1\over2u}\left({3\over4u^5} - {3\pi i\over2u^4} - {\pi^2\over u^3}\right)' + {i\pi\over u}\left({3\over4u^5} - {3\pi i\over2u^4} - {\pi^2\over u^3}\right)$$
$$ = {1\over2u}\left(-{15\over4u^6} + {6\pi i\over u^5} + {3\pi^2\over u^4}\right) + \left({3i\pi\over4u^6} + {3\pi^2\over2u^5} - {i\pi^3\over u^4}\right)$$
$$ = \left(-{15\over8u^7} + {15\pi i\over4u^6} +{3\pi^2\over u^5} - {\pi^3 i\over u^4}\right),\,\dots$$
One can obtain values of the function 
$$F(t) = {\sin2\pi\sqrt{t\,}\over\sqrt{t\,}} = \Im f(t)\tag3$$ 
and its derivations
$$F^{(n)}(t) = \Im f^{(n)}(t) = \Re g_n(\sqrt{t\,})\sin2\pi\sqrt{t\,} + \Im g_n(\sqrt{t\,})\cos2\pi\sqrt{t\,}\tag4$$
at $t=-1$. 
For example,
$$F(-1) = {\sinh(\pm2\pi i)\over\pm  i} = \sinh2\pi,$$
$$F'(-1) = \left.\left(-{1\over2t^{3/2}}\sin{2\pi\sqrt{t\,}} + {\pi\over t}\cos(2\pi\sqrt{t\,})\right)\right|_{t=-1}$$
$$= {\sin({\pm2\pi i})\over\pm2i} - \pi\cos({\pm2\pi i}) = {1\over2}\sinh2\pi - \pi\cosh2\pi,$$
$$F''(-1) = \left.\left(\left({3\over4t^{5/2}} - {\pi^2\over t^{3/2}}\right) \sin(2\pi\sqrt{t\,}) - {3\pi\over2t^2}\cos(2\pi\sqrt{t\,})\right)\right|_{t=-1}$$
$$ = \left({3\over4} + {\pi^2}\right)\sinh(2\pi) - {3\pi\over2}\cosh(2\pi),$$
$$F'''(-1) = \left.\left(\left(-{15\over8t^{7/2}} + {3\pi^2\over t^{5/2}}\right) \sin(2\pi\sqrt{t\,}) + \left({15\pi\over4t^3} - {\pi^3\over t^2}\right)\cos(2\pi\sqrt{t\,})\right)\right|_{t=-1}$$
$$= \left({15\over8} + {3\pi^2}\right)\sinh(2\pi) - \left({15\pi\over4} + \pi^3\right)\cosh(2\pi)\dots$$
The Taylor series in the neighborhood of the point $t=-1$ has the form
$$F(t) = {\sin2\pi\sqrt t\over\sqrt t} = {\sinh2\pi} + \sum_{n=1}^\infty c_n(t+1)^n,\tag5$$
where
$$c_n = {1\over n!}F^{(n)}(-1)\tag6$$
(see also Wolfram Alfa), and the series converges for $t\in\mathbb R.$
Substitution
$$t=z^2$$
gives the series in the form of
$${\sin2\pi z\over z} = {\sinh2\pi} + \sum_{n=1}^\infty c_n(z^2+1)^n,$$
so
$${\sin2\pi z\over z(z^2+1)} = {c_0\over z^2+1} + \sum_{n=1}^\infty c_n(z^2+1)^{n-1}$$
$$ = {1\over2i}{c_0\over(z-i)\left(1+\dfrac{z-i}{2i}\right)} + 
\sum_{n=0}^\infty c_{n+1}(z-i)^n(z-i+2i)^n$$
$$= c_0(z-i)^{-1}\sum_{n=0}^\infty(-1)^n{(z-i)^n\over(2i)^{n+1}}  + \sum_{n=0}^\infty c_{n+1}\sum_{k=0}^n\genfrac{(}{)}{0}{0}{n}{k}(2i)^{n-k}(z-i)^{n+k}.$$
Substitution $m=n+k\ $ in the double sum gives
$${\sin2\pi z\over z(z^2+1)}= c_0\sum_{n=0}^\infty(-1)^n{(z-i)^{n-1}\over(2i)^{n+1}}  + \sum_{m=0}^\infty \sum_{k=0}^\left[{m\over2}\right]c_{m-k+1}\genfrac{(}{)}{0}{0}{m-k}{k}(2i)^{m-2k}(z-i)^m$$
$$=  -{c_0i\over2}(z-i)^{-1}+ {c_0\over4}\sum_{m=0}^\infty\left({i\over2}\right)^m(z-i)^m  + \sum_{m=0}^\infty \left(i^m(z-i)^m \sum_{k=0}^\left[{m\over2}\right]i^{-2k}c_{m-k+1}\genfrac{(}{)}{0}{0}{m-k}{k}2^{m-2k}\right)$$
$$=  -{c_0i\over2}(z-i)^{-1}+ \sum_{m=0}^\infty \left(c_02^{-(m+2)} + \sum_{k=0}^\left[{m\over2}\right](-1)^k c_{m-k+1}\genfrac{(}{)}{0}{0}{m-k}{k}2^{m-2k}\right)i^m(z-i)^m$$
$$= -{c_0i\over2}(z-i)^{-1} + {c_0\over4} + c_1  + \left({c_0\over8} + 2c_2\right)i(z-i) - \left({c_0\over16} + 4c_3 - c_2\right)(z-i)^2 + \dots,$$
where
$$c_0 = \sinh2\pi,$$
$$c_1 = {1\over2}\sinh2\pi - \pi\cosh2\pi,$$
$$c_2 = \left({3\over8} + {\pi^2\over2}\right)\sinh2\pi - {3\pi\over4}\cosh2\pi,$$
$$c_3 = \left({5\over16} + {\pi^2\over2}\right)\sinh2\pi - \left({5\pi\over8} + {\pi^3\over6}\right)\cosh2\pi,\,\dots$$
I.e.,
$$\boxed{{\sin2\pi z\over z(z^2+1)} = - {i\sinh2\pi\over2}(z-i)^{-1}+ {3\over4}\sinh2\pi - \pi\cosh2\pi + i\left(\left({7\over8} + \pi^2\right)\sinh2\pi - {3\pi\over2}\cosh2\pi\right)(z-i) + \left(-\left({15\over16} + {3\pi^2\over2}\right)\sinh2\pi + \left({7\pi\over4}+{2\pi^3\over3}\right)\cosh2\pi\right)(z-i)^2 + \dots}$$