I am struggling to integrate $\frac{1}{(z-e^{-z})}$ on upper half unit circle. I would appreciate any support as I am very new to complex integration.
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1Is it the upper half or the right half? You say Re(z)>0... – Ian Dec 24 '17 at 19:33
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sorry, it is the upper half. I now corrected the question. – vyaman Dec 24 '17 at 19:56
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Is it just the circular arc or do you close it too? – Ian Dec 24 '17 at 20:53
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Only the circular arc – vyaman Dec 24 '17 at 21:29
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Since you're in the arc $|z| = 1$, and strictly speaking you're interested in the residue inside the unit circle, use Laurent Expansion.
$$\frac{1}{z - e^{-z}} = \frac{1}{e^{-z}(ze^z - 1)} = -\frac{e^z}{1 - ze^z} = -e^z\sum_{k = 0}^{+\infty}z^ke^{zk} = -\sum_{k = 0}^{+\infty}z^ke^{(k+1)z}$$
The series develops as follows
$$-1 - ze^{2z} - z^2e^{3z} - z^3e^{4z}-\ldots$$
By Cauchy formula, the residue is given by
$$2\pi i a_{-1}$$
Where $a_{-1}$ is the coefficient of $z^{-1}$ Laurent Series expansion, which here is missing. Hence the residue is zero and your integral is zero.
Enrico M.
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