complex analysis integration? I have to find $$\int_C \frac{2z-1}{z(z-1)} \,dz$$ where C is the circle |Z|=2
So,I thought about solving this by using Cauchy's formula.
That means I have  $f(x)=(2z-1)/z$
and then the integral = $2\pi\cdot i \cdot f(2)$ ...but how to continue this?
 A: I think you mean Cauchy. You can use the residue theorem - you have a first order pole at $ z = 0 $ and a first order pole at $ z = 1 $. The residue at $0$ is $$ \lim_{z\rightarrow0}(z - 0)\frac{2z - 1}{z(z-1)}=1 $$
and the residue at 1 is $$ \lim_{z\rightarrow1}(z - 1)\frac{2z - 1}{z(z-1)}=1 $$ so the contour integral is $2\pi i(1 + 1) = 4\pi i$.
A: Make little circles $\,C_0\,,\,C_1\,$around $\,0\;$ and $\;1\;$ resp.,  and connect them with the given circle $\,|z|=2\,$ , so that you now can apply Cauchy'Theorem on each little circle. Note that everywhere else the integral is zero as the function is analytic there:
$$\int\limits_{C_0}\frac{2z-1}{z(z-1)}dz=\left.2\pi i\left(\frac{2z-1}{z-1}\right)\right|_{z=0}=2\pi i\cdot 1=2\pi i $$
$$\int\limits_{C_1}\frac{2z-1}{z(z-1)}dz=\left.2\pi i\left(\frac{2z-1}z\right)\right|_{z=1}=2\pi i\cdot 1=2\pi i $$
Thus, the whole integral equals $\,2\pi i+2\pi i=4\pi i\;$ and the other answer, given by the method of residues, is correct and your book's wrong (nothing strange. Many books have mistakes)
A: Another easy approach is to use a partial fraction decomposition:
$$\frac{2z-1}{z(z-1)}=\frac{1}{z}+\frac{1}{z-1}$$
And apply Cauchy (or residue theorem) on both parts:
$$\implies \int_C \frac{2z-1}{z(z-1)} \,dz=2\pi i(1+1)=4\pi i$$
