Laurent series for $f (z)=\frac {\sin (2 \pi z)}{z (z^2 + 1)}$

Question

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

Answer 1

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.

Answer 2

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., $$\boxed{{\sin2\pi z\over z(z^2+1)} = - {i\sinh2\pi\over2}(z-i)^{-1}+ {3\over4}\sinh2\pi - \pi\cosh2\pi + i\left(\left({7\over8} + \pi^2\right)\sinh2\pi - {3\pi\over2}\cosh2\pi\right)(z-i) + \left(-\left({15\over16} + {3\pi^2\over2}\right)\sinh2\pi + \left({7\pi\over4}+{2\pi^3\over3}\right)\cosh2\pi\right)(z-i)^2 + \dots}$$