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I am having difficulty with with calculating the residue for $$\text{res}[\frac{\exp(\frac{1}{z})}{z^{2}-16,},z=0]$$ I was able to calculate the residues when $z=4$ and $z=-4$. However Im not sure how to approach this part of the question.

tired
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user6457350
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    Consider the integral around a circle of radius R centered at the origin. If R is greater than 4 then by the residue theorem, the integral is equal to 2pi times the sum of the residues. Now let R go to infinity. Try to compute that integral. – Clyde Nov 20 '16 at 17:17

2 Answers2

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Let $f(z) = \frac{e^{1/z}}{z^2 - 16}$. Let $C_R$ be the circle of radius R centered at the origin. For $R > 4$, $$ Res_{z=0}f(z) + Res_{z=4}f(z) + Res_{z = -4}f(z) = \frac{1}{2\pi i}\int_{C_R}f(z)dz $$

Note that $$ |\frac{1}{2\pi i}\int_{C_R} \frac{e^{1/z}}{z^2-16} dz| = |\frac{1}{2\pi i} \int_{0}^{2\pi} \frac{e^{(1/R) e^{-i\theta}}}{R^2e^{2i\theta} - 16}iRe^{i\theta} d\theta| \leq \frac{1}{2\pi}\int_0^{2\pi}\frac{R }{R^2 - 16}e^{(1/R)cos\theta} d\theta $$ $$ \leq \frac{1}{2\pi}\frac{R}{R^2 - 16} \int_{0}^{2\pi} e^{(1/R)} d\theta. $$

Letting $R \to \infty$, this integral goes to 0. Thus, $$ Res_{z=0}f(z) = - Res_{z=4}f(z) - Res_{z = -4}f(z). $$ Since you said you computed the right hand side, you now have the answer.

Clyde
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$e^z$ is an entire function, so $z = 0$ is an essential singularity for $e^{1/z}$.

$$e^{1/z} = \sum_{k = 0}^{+\infty} \frac{1}{k!}(z^{-k})$$

$$\frac{1}{z^2-16} = \frac{1}{16\left(\frac{z^2}{16} - 1\right)} = \frac{-1}{16} \sum_{n = 0}^{+\infty} \frac{z^{2n}}{16^n}$$

You can see the first terms of the expansion ad $z = 0$:

$$-\frac{1}{16}\left(1 + \frac{1}{z} + \frac{1}{2z^2} + \frac{1}{3!z^3} + \frac{1}{4!z^4} + \frac{1}{5!z^5} + \frac{1}{6!z^6} + \frac{1}{7!z^7} + \frac{1}{8!z^8} + \frac{1}{9!z^9} + \ldots \right) \times \\\\ \times \left(1 + \frac{z^2}{16} + \frac{z^4}{16^2} + \frac{z^6}{16^3} + \frac{z^8}{16^4} + \ldots\right)$$

Now you know that the residue is the coefficient of the term $z^{-1}$ and you can easily obtain it by finding all the term depending only on $z^{-1}$: for example the second terms on the left times the first term on the right bracket, plus the the fourth term on the left bracket times the second on the right bracket and so on, and you will get

$$-\frac{1}{16}\left(\frac{1}{z} + \frac{1}{16\times 3! z} + \frac{1}{16^2\times 5! z} + \frac{1}{16^3\times 7! z} + \ldots\right)$$

It's easy to guess the series for that expansion:

$$\frac{-1}{16z} \sum_{k = 0}^{+\infty} \frac{1}{16^k\cdot (2k + 1)!}$$

The series converges to

$$4\sinh \left(\frac{1}{4}\right)$$

Hence the residue is

$$\frac{-1}{4} \sinh \left(\frac{1}{4}\right)$$

That is indeed the value of your residue

$$-0.06315307920204207...$$

Enrico M.
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  • there is a residue at $z=0$ which can be deduced as the $a_{-1}$ coefficent of the expansion you give above – tired Nov 20 '16 at 17:11
  • @tired mmm I surely made some error... – Enrico M. Nov 20 '16 at 17:12
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    Hint: all terms of the above double sum which fullfil that $-k+2n=1$ will contribute to the residue. can you figure out its closed form (not too complicated). – tired Nov 20 '16 at 17:17
  • @tired You're right, I managed to find that the residues is $95/96$. But there has to be some error.. I mean the residue shalln't exist. Or does it? I'm confused. – Enrico M. Nov 20 '16 at 17:22
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  • the residue exists. the definition of a residue as a specific coefficent in a laurent expansion holds regardless of the exact nature of the singularity
  • 9/96 is wrong... i think the result should be $\approx -0.0631531 i$
    • – tired Nov 20 '16 at 17:25
  • @Sorry, it was a typo: I meant $95/96$. What confuses me is that if I try with Mathematica, it says nothing about. But by hands it's $95/96$ (or in any case, it does exist, wether I'm right or wrong about the numerical value) – Enrico M. Nov 20 '16 at 17:26
  • mathematica fails for this kind of residues. and i still think that your numerical value is false – tired Nov 20 '16 at 17:32
  • @tired I forgot a $-16$ factor :D But even with that, the residue I found is different from yours. mm.. I have to check it again.. – Enrico M. Nov 20 '16 at 17:34
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    the solution is $-i \sinh(1/4)/4$...you will do it! – tired Nov 20 '16 at 17:39
  • @tired I checked the clever solution above! Your result is correct (of course). I will work on this then! – Enrico M. Nov 20 '16 at 17:40
  • a last hint: the resdiue is an infinite sum of terms which might be given in a simple closed form – tired Nov 20 '16 at 17:41
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    @Tired Got it!! Going to edit! – Enrico M. Nov 20 '16 at 17:59
  • still false...:(# – tired Nov 20 '16 at 18:39
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    @tired I forgot a factorial, lol. – Enrico M. Nov 20 '16 at 18:44
  • @tired P.s. Thank you for having guided me to that! I learnt something ^_^ – Enrico M. Nov 20 '16 at 18:45
  • now you have to find the $i$. but for your nice effort i give you (+1) now – tired Nov 20 '16 at 18:46
  • @tired Oh Gosh, right, the $i$... – Enrico M. Nov 20 '16 at 18:54