My attempt
$$\int_1^{-1} \frac{1}{x+i} \, dx+\int_1^{-1} \frac{i}{-1+i y} \, dy+\int_{-1}^1 \frac{1}{x-i} \, dx+\int_{-1}^1 \frac{i}{1+i y} \, dy=2\pi i$$
Is this correct?
How to integrate $\frac{1}{z}$ around square with vertices $(1,1),(-1,1),(1,-1),(-1,-1)$?
2
$\begingroup$
integration
complex-analysis
ordinary-differential-equations
2 Answers
1
This is correct. Since $\frac{1}{z}$ ist holomorphic except for the origin, you can change the path, as long you don't mix up with that singularity.
You have essantially a (counter-clockwise) circle around the origin, where a pole of order one is located.
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0Where do you mean "$\,\frac1z\;$ is holomorphic "...? And if it is holomorphic, what singularities are you talking about? – 2017-01-06
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0Sorry, complex analysis is some years in the past. I meant "holomorphic everywhere but the origin", so the path can be changed. I'll edit the anwser. – 2017-01-06
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0Perhaps "analytic except at..." is more standard, but it doesn't really matters once you mention the singularities. – 2017-01-06
0
It's correct, but you don't have to calculate any integrals anyway. This is the core of the Residue theorem: an integral over a closed loop equals the sum of the encircled residues times $2\pi i$ (factors in front of poles of order $1$). In this case you have a single $1/z$ pole at the origin ($z=0$) with a prefactor of $1$, so the result is $2\pi i$ without looking.