Cauchy's Residue Theorem 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. 
 A: 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 $\mbox{Res}(3/(z-1)) = 3$.
A: 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.
A: 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 $\,10\pi i$
