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The function $\displaystyle g(z) = \frac{e^{iz}-1}{\cos z-1}$ has a Laurent expansion of the form $\sum_{n=-\infty}^{+\infty} c_{n}z^{n}$ in the region $2\pi<|z|<4\pi$. Find the coefficient $c_{-3}$.

I am not sure how to proceed with problems of this sort, the only method I know to find Laurent series is to manipulate geometric series. I assume in this case you have to do residue calculations? I would appreciate any input very much, I am studying before an exam in complex analysis.

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  • $\begingroup$ You can use plain old high-school division, analogous to the long division of polynomials, but writing terms in greater degree to the right. You know the expansion of top and of bottom, then just divide. $\endgroup$ – Lubin Aug 24 '13 at 17:28
  • $\begingroup$ @Lubin look at the region of convergence - your method cannot be used $\endgroup$ – user8268 Aug 24 '13 at 17:45
  • $\begingroup$ @Lubin, why can you use this method to find the coefficient $c_{3}$ but not $c_{-3}$? It seems hard to find the expansions of the top and the bottom under this region of convergence. $\endgroup$ – Sid Aug 24 '13 at 17:52
  • $\begingroup$ Sorry, I have seriously misread the problem. $\endgroup$ – Lubin Aug 24 '13 at 18:08
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(to show that it is a problem you can expect for the exam)

$c_{-3}=\frac{1}{2\pi i}\oint_{z=c} z^2\frac{e^{iz}-1}{\cos z -1} dz$ for any $2\pi<c<4\pi$. The integrant has possible singularities when $\cos z=1$, i.e. for $z=2\pi n$. Inside the circle these are $z=0,2\pi,-2\pi$. We thus have $$c_{-3}=\sum_{w\in\{0,2\pi,-2\pi\}}Res_{w}z^2\frac{e^{iz}-1}{\cos z -1}$$ (by the residue theorem}. To compute the residues (use any method you know/like, this is one): $e^{i(x+w)}-1=e^{ix}-1=ix+\dots$, $\cos(x+w)-1=\cos x -1=-x^2/2+\dots$ (for any $w=2\pi n$), hence the residue is $w^2\frac{i}{-1/2}=-2iw^2$. Their sum is $-4i(2\pi)^2=-16i\pi^2$.

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  • $\begingroup$ Thanks, this seems like the most practical and methodological solution to problems of this sort. I am indebted to you guys. $\endgroup$ – Sid Aug 24 '13 at 19:00
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Writing $e^{iz} = \cos z + i\sin z$, we obtain

$$\begin{align} \frac{e^{iz}-1}{\cos z - 1} &= 1 + i \frac{\sin z}{\cos z - 1}\\ &= 1 + i \frac{2\sin (z/2)\cos(z/2)}{-2\sin^2 (z/2)}\\ &= 1 - i \cot (z/2). \end{align}$$

Now, for the cotangent, we have the partial fraction decomposition

$$\pi \cot (\pi w) = \frac1w + \sum_{\nu = 1}^\infty \frac{2w}{w^2-\nu^2},$$

from which we obtain

$$\cot (z/2) = \frac2z + \sum_{\nu=1}^\infty \frac{4z}{z^2 - (2\pi\nu)^2}.$$

The only terms with singularities in $\{\lvert z\rvert \leqslant 2\pi \}$ are $\frac2z$ and the term for $\nu = 1$,

$$\frac{4z}{z^2 - (2\pi)^2} = \frac{4}{z}\cdot \frac{1}{1 - \left(\frac{2\pi}{z}\right)^2} = \frac{4}{z}\sum_{\kappa=0}^\infty \left(\frac{2\pi}{z}\right)^{2\kappa}$$

and the coefficient of $z^{-3}$ in that is $4(2\pi)^2 = 16\pi^2$, whence the coefficient $c_{-3}$ in the Laurent series of

$$ \frac{e^{iz}-1}{\cos z - 1}$$

in the annulus $2\pi < \lvert z\rvert < 4\pi$ is $-i16\pi^2$.

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  • $\begingroup$ After a false start, this works. $\endgroup$ – Daniel Fischer Aug 24 '13 at 18:18
  • $\begingroup$ Thisfinal answer matches the one in my solution manual, so this must be correct. Thank you very much. Apparently this problem is a lot more complicated than I assumed, I did not expect to need to memorize this particular fraction decomposition for cotangent on my exam. $\endgroup$ – Sid Aug 24 '13 at 18:25
  • $\begingroup$ Just because it matches doesn't necessarily mean it is correct. However, since it matches, I'm pretty confident that I didn't make any mistakes when rescaling the cotangent partial fraction. Everything else is pretty simple. You don't need the partial fraction decomposition, it is sufficient to find the residues in $0, \pm 2\pi$, then you have - since the poles are simple - three principal parts $\frac{a}{z}$, $\frac{b}{z-2\pi}$ and $\frac{c}{z+2\pi}$, then develop these - well, the two nontrivial ones - into Laurent series in the annulus and add the coefficients. $\endgroup$ – Daniel Fischer Aug 24 '13 at 18:31
  • $\begingroup$ The nice thing about using the partial fraction of the cotangent is that it makes it very easy to find the principal part of the Laurent series in other annuli. $\endgroup$ – Daniel Fischer Aug 24 '13 at 18:32
  • $\begingroup$ Thanks! I'm attempting to solve the problem right now using this method. $\endgroup$ – Sid Aug 24 '13 at 18:58

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