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Are there some conditions under which the following two are equal?

$$\displaystyle \oint_C \sum f_n(z)= \sum \oint_C f_n(z)$$

In the case of real valued functions, the condition $f_n(z) \geq 0$ suffices (see here). Is there a simple condition like that in the case of complex valued analytic functions with simple/double poles?

Thank you :)

Community
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Isomorphism
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  • An analogous condition for complex-valued functions would be the requirement that, for $z \in C$, the functions $f_n(z)$ take values in one, and only one, quadrant. – Antonio Vargas Aug 20 '14 at 19:39
  • Sigh! I want a condition that allows me to run a contour integral around a circle that is centered at origin :( – Isomorphism Aug 21 '14 at 01:14
  • Sure, that's fine, what I said still applies. I've clarified my comment in an answer. – Antonio Vargas Aug 21 '14 at 01:34

2 Answers2

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Since

$$ \int_C f(z)\,dz = \int_C \operatorname{Re} f(z)\,dz + i\int_C \operatorname{Im} f(z)\,dz, $$

you can essentially apply the same criteria in the complex case as in the real case. Specifically, you just need the real and imaginary parts of the integrands to be single-signed.

Let $Q_j$, $j=1,2,3,4$, denote the four closed quadrants of the complex plane. If

$$ f_n(z) \in Q_j \quad \text{for all } n \in \mathbb N \text{ and } z \in C $$

for some $Q_j$ then

$$ \displaystyle \oint_C \sum_n f_n(z)= \sum_n \oint_C f_n(z) $$

follows from applying the known result for real functions to the real and imaginary parts of the integral.

Antonio Vargas
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Uniform convergence allows you to switch a limit and an integral. So if the $f_n$ converge uniformly to $0$ on $C$ (for instance, if they are the terms of a Laurent series and $C$ is contained inside the annulus of convergence), then $\sum f_n(z)$ converges uniformly on $C$, and it is safe to switch the sum and the integral.

Brett Frankel
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  • How would you check if $f_{n}$ converges uniformly, also would the result work if we are integrating $c^{1}$ curves ? – Zophikel Sep 11 '17 at 20:17