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I have to compute the contour integral $\int_{\gamma} \frac{(z-1)}{z(z+1)(z-2)}dz$ where $\gamma$ is (a) the circle $C(0;1)$, (b) the circle $C(0,\frac{3}{2})$ and (c) the rectangle of vertex $-2-i$, $-2+i$, $3+i$ and $3-i$. I am not used to with that kind of integral.

For (b), I obtain $$\int_{\gamma} \frac{(z-1)}{z(z+1)(z-2)}dz = \int_{\gamma} \frac{1}{2z}dz - \int_{\gamma} \frac{2}{3(z+1)}dz+ \int_{\gamma} \frac{1}{6(z-2)}dz = (\frac{1}{2}-\frac{2}{3}+\frac{1}{6}) 2 \pi i =0$$ by Cauchy theorem.

Someone explain to me since the point $z=2$ is out of the circle, so the contour integral $\int_{\gamma} \frac{1}{(z-2)}dz$ is equal to zero. I think this explication is related to the holomorphy of $\frac{1}{z-2}$, but it is very unclear for me. Is anyone could explain to me rigorously why this fact is true?

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  • $\begingroup$ $\;z=-1\;$ is a singularity of the integrand on $\;C(0,1)\;$ . How do you intend to do this integral in (a) ? $\endgroup$ – DonAntonio Mar 13 '16 at 15:11
  • $\begingroup$ This is for the (a), but in this situation I'll compute $\int_{\gamma} \frac{dz}{z+1}$ in using the parametrization $z(t)=\cos t +i \sin t$ with $-\pi \leq t < \pi$ and the two other contour integrals in using the Cauchy theorem. However, my question is on the part (b) $\endgroup$ – user1050421 Mar 13 '16 at 15:15
  • $\begingroup$ @Joanpemo Does the idea you have in mind? $\endgroup$ – user1050421 Mar 13 '16 at 15:26
  • $\begingroup$ About your question's last part: since the domain enclosed by $\;\gamma\;$ is simply connected and $\;\frac1{z-2}\;$ analytic there, Cauchy's Theorem gives zero as the integral's value. $\endgroup$ – DonAntonio Mar 13 '16 at 15:31
  • $\begingroup$ Which integral? The main integral or $\int_{\gamma} \frac{dz}{z-2}$? $\endgroup$ – user1050421 Mar 13 '16 at 15:37
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Since $\;0,\,-1\in C_1:=C\left(0,\,\frac32\right)\;$ and these are simple poles, we have that

$$\oint_{C_1}\frac{z-1}{z(z+1)(z-2)}dz=2\pi i\left(\text{Res}_{z=0}(f)+\text{Res}_{z=-1}(f)\right)=$$

$$=2\pi i\left(\lim_{z\to0}\,zf(z)+\lim_{z\to-1}(z+1)f(z)\right)=2\pi i\left(\frac12-\frac23\right)=-\frac{\pi i}3$$

Another way: draw small circles $\;C_0,\,C_{-1}\;$ ( say, of radius $\;0.05\;$) around each of $\;z= 0\,,\,\,z=-1\;$ respectively. Then the integral on $\;C_1\;$ equals the sum of the integrals over these small circles (join the big circle with a small one by a straight segment and etc.), and

$$\oint_{C_1}f(z)dz=\oint_{C_0}\frac{\frac{z-1}{(z+1)(z-2)}}zdz+\oint_{C_{-1}}\frac{\frac{z-1}{z(z-2)}}{(z+1)}dz=$$

$$=2\pi i\left(\left.\frac{z-1}{(z+1)(z-2)}\right|_{z=0}+\left.\frac{z-1}{z(z-2)}\right|_{-1}\right)=2\pi i\left(\frac12-\frac23\right)=-\frac{\pi i}3$$

Using Cauchy's Formula

$$f(z_0)=\frac1{2\pi i}\oint_C\frac{f(z)}{z-z_0}dz$$

under the usual conditions.

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  • $\begingroup$ I can't use the Residue theorem, because I have not seen that in class yet. Could you modify your answer in such a way I could well understand? :) $\endgroup$ – user1050421 Mar 13 '16 at 15:32
  • $\begingroup$ @RobertDavis Check now. If this also isn't within the material you've already studied I'm afraid you'll have to do it the ugly, nasty way: with parametrization, line integrals and etc. $\endgroup$ – DonAntonio Mar 13 '16 at 15:44
  • $\begingroup$ All your answer is very good except the last line; It miss $\frac{1}{2 \pi i}$. $\endgroup$ – user1050421 Mar 13 '16 at 17:43
  • $\begingroup$ @RobertDavis Thank you very much. Forgot it...hehe. Edited now. $\endgroup$ – DonAntonio Mar 13 '16 at 17:44

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