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Compute $$\oint \left[ z e^{3/z} + \frac{\cos z}{z^2 (z - \pi )^3} \right] \, dz$$ $$|z| = 5$$

My question is how to do residue at $$\oint ze^{3/z} \, dz $$

dustin
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razercreed
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    Have you tried expanding $e^{3/z}$ using a Laurent series? – JimmyK4542 Jan 02 '15 at 02:43
  • @JimmyK4542 no did i need to do it can u explain why? – razercreed Jan 02 '15 at 02:49
  • As $z\to0$, then $w=3/z\to\infty$, so think of how $e^w$ behaves as $w\to\infty$. Depending on whether $w$ moves along the real axis in the positive direction, or in the negative direction, or along the imaginary axis, $e^w$ may approach $\infty$ or $0$ or may oscillate. So you have an essential singularity rather than a pole. ${}\qquad{}$ – Michael Hardy Jan 02 '15 at 02:52

2 Answers2

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Hint:

As @JimmyK4542 suggested, let's expand $e^{3/z}$ using a Laurent series: $$ e^z = \sum_{n=0}^{\infty}\frac{z^n}{n!} $$ We can plug in $3/z$ and get: $$ e^{3/z} = \sum_{n=0}^{\infty}\frac{(\frac{3}{z})^n}{n!} = \sum_{n=0}^{\infty}\frac{3^n}{z^nn!} = 1 + \frac{3}{z} + \frac{9}{2z^2} + \dots $$ From here we get: $$ ze^{3/z} = z + 3 + \frac{9}{2z} + \dots $$

What is the residue of $ze^{3/z}$ then?

benji
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You have two poles inside your contour. Namely $z=0$ which is an essential singularity and $z=\pi$ which is a pole of order $3$. For the first pole $z=0$ we will have the Laurent expansion around the point $z=0$

$$ z e^{3/z} + \frac{\cos z}{z^2 (z - \pi )^3} = z\left(1+\frac{3}{z}+ \frac{9}{2!\,z^2}+\dots \right) + \left(\frac{1}{2\pi^3}-\frac{6}{\pi^5} - {\frac {3}{{\pi }^{4}z}}-{\frac {1}{{z}^{2}{\pi }^{3}}}+\dots \right). $$

I think you should be able to find the residue from the above Laurent series which corresponds to the pole $z=0$.

I let you finish the problem. See related problems and techniques.

Community
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Mhenni Benghorbal
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