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I'm having trouble evaluating the following contour integral: $$I=\oint \frac{z^3e^{\frac{1}{z}}}{1+z^3}dz\ ;\ \ \ \ \ \ \text{for the anticlockwise circle defined by: }\quad |z|=3$$

What I've done is to realice that all singularities, including the essential one, fall within the circle:

My first step was to factorize de denominator like so: $$1+z^3=\left(z + 1\right) \left(z-\frac{1+ \sqrt{3}\ i}{2}\right)\left(z-\frac{1-\sqrt{3}\ i}{2}\right)$$

since the numerator is well defined in all the solutions of the above polinomial, I can evaluate the residues without problem, evaluating the numerator where needed. This residues are $R_1, R_2$ and $R_3$

For the essential singularity this was what I did:

Since the problem is un the numerator, I wrote down a Laurent series for it...

$$z^3e^{\frac{1}{z}}=z^3\sum_{n=0}^{\infty}\frac{1}{n!z^n}=\sum_{n=0}^{\infty}\frac{z^{3-n}}{n!}$$

Here is the part that apparently is wrong:

I considered tha the residues in this point is the factor in the series for which the exponent equals -1, since the denominator of the integral is 1 in $z=0$, this residue which I will name $R_4$ is:

$$R_4=\frac{1}{4!}$$

Hence the integral is:

$$I=2\pi i(R_1+R_2+R_3+R_4)$$

Does anyone know if this is the correct solution?

Thank you all in advance.

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An easy way is to substitute $z=\frac{1}{w}$ so that $dz=-\frac{1}{w^2}$ with contour $C$:|$w$|$=\frac{1}{3}$ clockwise and your integral is $$\oint_C\frac{-e^w}{w^2(1+w^3)}dw$$.

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  • $\begingroup$ But that mapping takes everything that was inside the circle and put it outside. Or not? $\endgroup$ – mytorojas Jul 25 '19 at 14:32
  • $\begingroup$ yes and that helps you as calculation of residue is required at $w=0$ only. $\endgroup$ – Nitin Uniyal Jul 25 '19 at 16:21
  • $\begingroup$ In a sense you are doing the same that @RobertZ did, you are bringing $\infty \to 0$? $\endgroup$ – mytorojas Jul 25 '19 at 17:08
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    $\begingroup$ Other poles viz. $-1,-\omega,-\omega^2$ will lie outside $C$. $\endgroup$ – Nitin Uniyal Jul 26 '19 at 3:35
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    $\begingroup$ @mytorojas Substituting $w=1/z$ and then finding residue at pole $w=0$ is just THE SAME as considering the residue at infinity ($z=\infty\leftrightarrow w=1/z=0$) $\endgroup$ – Robert Z Oct 1 '19 at 5:19
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Yes, $I=2\pi i(R_1+R_2+R_3+R_4)$, but $R_4$, your residue evaluation at $z=0$ is wrong. It should be the coefficient of $1/z$ in the product $$\frac{z^3}{1+z^3}\cdot e^{1/z} =\sum_{n=0}^{\infty}(-1)^n z^{3n}\cdot\sum_{n=0}^{\infty}\frac{1}{n!z^n}$$ that is $$\frac{1}{4!}-\frac{1}{7!}+\frac{1}{10!}-\frac{1}{13!}+\dots.$$

Since the function to integrate is holomorphic outside the circle $|z|=3$, it is easier to consider the residue at infinity, $$I=\oint_{|z|=3} \frac{z^3e^{\frac{1}{z}}}{1+z^3}dz=2\pi i\cdot\text{Res}\left(\frac{(1/z)^3e^{z}}{z^2(1+1/z^3)},0\right)=2\pi i\cdot\text{Res}\left(\frac{e^{z}}{z^2(z^3+1)},0\right)=2\pi i.$$

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  • $\begingroup$ @mytorojas Any further doubt? $\endgroup$ – Robert Z Jul 25 '19 at 6:27
  • $\begingroup$ No, thank you very much $\endgroup$ – mytorojas Jul 25 '19 at 6:29
  • $\begingroup$ I have a doubt, the $I$ you wrote, is the whole integral or just the contribution of the essential singularity? $\endgroup$ – mytorojas Jul 25 '19 at 16:06
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    $\begingroup$ $I$ is the whole integral, the same notation of yours. $\endgroup$ – Robert Z Jul 25 '19 at 16:08
  • $\begingroup$ Thak you very much! $\endgroup$ – mytorojas Jul 25 '19 at 16:09

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