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I need to evaluate $\int_C \frac{5z-2}{z(z-1)}dz$ where $C$ is the circle $|z|$=2. I used partial fraction decomposition to get $$\frac{5z-2}{z(z-1)}=\frac{2}{z}+\frac{3}{z-1}$$ I have the answer ($10\pi i$) in my book but I don't fully understand how to get it. I think I have to evaluate in the two domains $0<|z|<1$ and $0<|z-1|<1$, is it because the singularities are at $z=0$ and $z=1$? I ask mainly because of the $|z-1|$ in the second inequality.

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    $\begingroup$ [spam]$(2+3)\cdot 2\pi i = 10 \pi i$ :)[/spam] $\endgroup$ – Cortizol Apr 4 '13 at 18:17
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I don't think you need the Residue Theorem, but just Cauchy's integral formula? Note that $$\dfrac{5z - 2}{z(z-1)} = \frac{3}{z-1} + \frac{2}{z}$$ and so $$\int_{C} \frac{5z-2}{z(z-1)}dz = \int_{C}\frac{3}{z-1} dz + \int_{C}\frac{2}{z} dz = 2\pi i f_1(1) + 2\pi i f_2(0)$$ where $f_1(z) = 3$ and $f_2(z) = 2$, so $2\pi i f_1(1) + 2\pi i f_2(0) = 2\pi i \cdot 3 + 2\pi i \cdot 2 = 10 \pi i$.

However, you can also use the residue theorem. Note that $\frac{3}{z-1}$ is already a Laurent series expanded about $z = 1$ when $0 < |z-1| < 1$ and $\frac{2}{z}$ is one as well expanded about $z = 0$ when $0 < |z| < 1$. Therefore, $$\dfrac{5z - 2}{z(z-1)} = \frac{3}{z-1} + \frac{2}{z}$$ and so $$\int_{C} \frac{5z-2}{z(z-1)}dz = \int_{C}\frac{3}{z-1} dz + \int_{C}\frac{2}{z} dz = 2\pi i \mbox{Res}\left(\frac{3}{z-1}\right) + 2\pi i\mbox{Res}\left(\frac{2}{z}\right)$$ where $$\mbox{Res}\left(\frac{3}{z-1}\right) = 3 \quad \mbox{and} \quad \mbox{Res}\left(\frac{2}{z}\right) = 2$$ so you again get $10 \pi i$.

The choice of the domains is because $0 < |z| < 1$ and $0 < |z-1| < 1$ is where the function $(5z-2)/(z(z-1))$ is analytic and so justifies the residues. That is, $2/z$ is already a Laurent series when $0 < |z| < 1$ with $\mbox{Res}(2/z) = 2$ and $3/(z-1)$ is already a Laurent series when $0 < |z-1| < 1$ with $\mbox{Res}(3/(z-1)) = 3$.

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  • $\begingroup$ Thank you, but how do they get the domains $0<|z|<1$ and $0<|z-1|<1$? $\endgroup$ – Alti Apr 4 '13 at 18:40
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    $\begingroup$ Hi Alti, please see the added line above. Let me know if you need further explanation. $\endgroup$ – Suugaku Apr 4 '13 at 19:14
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For a simple pole $z_0$ of a function $f(z)$, a residue at that pole is

$$\lim_{z \rightarrow z_0} [(z-z_0) f(z)]$$

You do not need to do partial fractions to evaluate the sum of the residues. For a simple pole, you can effectively ignore the piece that is blowing up. For your example, the residue at $z=0$ is

$$\frac{5 \cdot 0 -2}{0-1} = 2$$

and the residue at $z=1$ is

$$\frac{5 \cdot 1 -2}{1} = 3$$

The sum of the residues is then $5$. Note that these are both within $C$, so the whole sum contributes to the integral.

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  • $\begingroup$ Thank you. I haven't gotten to poles yet, this section is slightly before it, so that's why I used partial fraction. $\endgroup$ – Alti Apr 4 '13 at 18:21
  • $\begingroup$ But notice that you are doing precisely what I have done anyway to evaluate the residues. You multiply the first fraction by $z$ and get $2$, and you multiply the 2nd fraction by $z-1$ and get $3$. $\endgroup$ – Ron Gordon Apr 4 '13 at 18:22
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Hints:

According to Cauchy's Formula, for a nice function (and your function is nice...why?) we have

$$f(z_0)=\frac{1}{2\pi i}\int\limits_C\frac{f(z)}{z-z_0}dz$$

So take two little circles: one ($\,C_0\,$) around $\,z=0\,$ and another ($\,C_1\,$) around $\,z=1\,$ , and join $\,|z|=2\,$ with them by to straight lines (how?), so:

$$\int\limits_{C_1}\frac{\frac{5z-2}{z-1}}{z}dz=\left.2\pi i\left(\frac{5z-2}{z-1}\right)\right|_{z=0}=4\pi i$$

$$\int\limits_{C_1}\frac{\frac{5z-2}{z}}{z-1}dz=\left.2\pi i\left(\frac{5z-2}{z}\right)\right|_{z=1}=6\pi i$$

Add the above (justification?) and get indeed $\,10\pi i$

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  • $\begingroup$ Interesting downvote: no explanation...oh,well. Something wrong? Something missing? Something misunderstood...? $\endgroup$ – DonAntonio Apr 4 '13 at 18:40
  • $\begingroup$ I was confused about that as well. Thank you for your help. $\endgroup$ – Alti Apr 4 '13 at 18:42
  • $\begingroup$ Anytime, @Alti.... $\endgroup$ – DonAntonio Apr 4 '13 at 18:43

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