I would like to find the poles and residues of $$ f(z) = \frac {1}{z^2 \sin(\pi (z + \alpha))} $$ Where $0<\alpha<1$.

I found the first pole, which is a double pole at $z=0$, and then also there are poles at $z=n-\alpha$.

To find the residues I used the Laurent expansion, so for the residue of $z_0 = n - \alpha$

$$ f(z) = \frac {a_{-1}}{z-z_0} + a_0 + a_1(z-z_0) + ... $$

I multiplied $f(z)$ by $z-z_0$ and then took the limit as $ z \rightarrow z_0$

I don't understand how to then find $a_{-1}$, because I'm not sure what to do after this point: $$ \displaystyle \lim_{z\rightarrow z_0} \left[\frac{z-(n-\alpha)}{(n-\alpha)^2 \sin(\pi(z+\alpha))}\right]=a_{-1} $$

I'd appreciate guidance through the method of finding $a_{-1}$ and also what method to use for the $z^2$ pole.


The poles at $\;z=n-\alpha\;,\;\;n\in\Bbb Z\;$ are simple , so

$$Res_{z=n-\alpha}(f)=\lim_{z\to n-\alpha}\frac{z-(n-\alpha)}{z^2\sin(\pi(z+\alpha))}=\lim_{z\to n-\alpha}\frac1{2z\sin(\pi(z+\alpha)+\pi z^2\cos(\pi(z+\alpha))}=$$$${}$$

$$=\frac1{\pi(n-\alpha)^2\cos\pi n}=\frac{(-1)^n}{\pi(n-\alpha)^2}$$

  • $\begingroup$ To find residue of z^2 pole, would i replace $z-(n-\alpha)$ with $z-0$ to find the residue of $z^2$? Then the z in the numerator and denominator would cancel $\endgroup$ – inya Jun 12 '17 at 17:36
  • 1
    $\begingroup$ @inya The pole at $z=0$ is of order $2$, so you have to use $$ res_{z=0} (f) = \displaystyle \lim_{z\to 0} \frac{d}{dz} \left(\frac{z^2}{z^2 \sin(\pi(z+\alpha))}\right)$$ $\endgroup$ – bob Jun 12 '17 at 17:42

For simple poles, a lot of the time computing the Laurent series isn't the easiest way to do it here are a few tricks:

1) if $f=\frac{h}{g}$ where $h,g$ are holomorphic has a pole at $z=a$

then $Res_{z=a}(f) = \frac{h(a)}{g'(a)}$

2) or in general, $Res_{z=a}(f) = \lim_{z->a}{(z-a)f(z)}$

(note these only holds for simple poles)

So you can use the first trick to find the residues of $n-\alpha$:

$Res_{z=n-\alpha}(f) = 1/(\frac{d}{dz}z^2sin(π(z+α))) |_{z=n-\alpha}$

$=\frac{1}{2zsin(\pi(z+\alpha) + \pi z^2cos(\pi (z-\alpha)))} |_{z=n-\alpha}$

$=\frac{(-1)^n}{\pi (n-\alpha)^2}$

You can also use trick 2 as you did, to get that to work you can use L'Hopital's rule (as DonAntonio did)

Now for the double pole at 0, finding the Laurent series of $sin(\pi (z+ \alpha))$ or $cosec(\pi (z+ \alpha))$ about 0 doesn't look easy so in this case the easiest way might(?) be to use the more general form of trick 2: $Res_{z=a}(f) = \frac{1}{(m-1)!}\lim_{z->a}{\frac{d^{m-1}}{dz^(m-1)}(z-a)^m f(z)}$ , for a pole of order m

so in this case:

$Res_{z=0}(f) = \frac{d}{dz}cosec(\pi (z+\alpha))|_{z=0}$



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