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?
Best Answer
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.