I'm calculating some countour integrals of complex functions by residues, but I can't understand an certain way to calculate these residues in some cases.

We have a theorem that states that the integral of a function $f$ over a closed simple contour $C$ is given by the sum of the residues inside that contour. By definition, I'm calculating the residue of a function $f$ in a singular point $z_0$ by finding the Laurent series of $f$ around that point and taking the coefficient of the $\frac{1}{z-z_0}$ term.

But in a example given in my text, they calculate the residues of $\frac{5z-2}{z(z-1)}$ by writing it in the partial fractions form $$ \frac{2}{z} + \frac{3}{z-1}, $$ and then the residues in $z = 0$ and $z = 1$ are claimed to be $2$ and $3$, respectively, because "$2/z$ and $3/(z-1)$ are already Laurent series around the the respective points". I see they indeed are, but their sum are not (correct me here if I'm wrong) a Laurent series of $f$ around one of the points, so I don't see how they can each give either of the residues.

To ensure that my doubt will be understood, that's the way of finding the residues I do understand: for the same function we can write it, for $0<|z|<1$, as $$ f(z) = \frac{5z-2}{z}\cdot\frac{-1}{1-z} = \left(5 - \frac{2}{z}\right)(-1 -z -z^2 - \cdots); $$ so that multiplying the term at left to the series in the right member we easily get the coefficient of $1/z$, thus finding the residue at $z = 0$; and a analogous procedure can be used to find residue at $z = 1$. But in both of these we write $f$ as a series of powers of $z-z_0$ that is a Laurent series around the needed point, by definition. I'm struggling to see why the partial fractions quite do the same.

Thanks in advance.

  • 3
    $\begingroup$ $3/(z-1)$ is analytic at $z=0$, so it can't contribute to the residue there whatever its expansion. Likewise $2/z$ is analytic at $z=1$ and has no bearing on the residue at $1$. $\endgroup$ – sharding4 Jul 1 '18 at 1:06
  • $\begingroup$ @sharding4 Sorry, I thought a lot about what you said but I don't understand what you mean with "contribuiting to the residue". It seems to me that you're using an alternative statement about calculation of residues that I do not know. $\endgroup$ – AnalyticHarmony Jul 1 '18 at 4:12
  • $\begingroup$ What I mean is, yes, $3/(z-1)$ is analytic at $z = 0$ but I'm only using the definition of residue that it is the coefficient of the $(1/z)$ term of the Laurent series, so that doens't say nothing to me about the value of the residue of $f$ at that point. $\endgroup$ – AnalyticHarmony Jul 1 '18 at 4:13
  • 2
    $\begingroup$ If you expand $3/(z-1)$ about $z=0$ you'll get a Taylor's series with only non-negative powers of the variable, because it's an analytic function at $z=0$. With no $1/z$ term the expansion of $3/(z-1)$ can't affect the residue (the coefficient of the $1/z$ term) It's just a geometric series. Write out its expansion and check. $\endgroup$ – sharding4 Jul 1 '18 at 4:21
  • $\begingroup$ Your series for $1/1-z$ does not converge in a neighborhood of $0$. So you cannot integrate the expression on the right term by term when the contour encloses $0$ in its interior. However, you can integrate the partial fraction expression over such a contour. $\endgroup$ – DisintegratingByParts Jul 2 '18 at 0:10

If you want to compute $\oint_{C}f(z)dz$ where there are a finite number of singularities $z_1,z_2,\cdots,z_N$ inside $C$ and none on $C$, then you can write this as $N$ integrals, each one enclosing and encircling just one of the points $z_k$. That one integral around $z_k$ can be computed by expanding in a Laurent series about $z_k$, and looking at the coefficient of the $1/(z-z_k)$ term. Alternatively, if the function $f$ has a partial fraction expansion, then the integral around $z_k$ will be $0$ for all terms not involving a term $1/(z-z_k)$.


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