Evaluating $\int_C\frac{z+1}{z^2-2z}dz$, where $C$ is the circle $|z|=3$ 
Evaluate the contour integral $\int_C\frac{z+1}{z^2-2z}dz$ using Cauchy's residue theorem, where $C$ is the circle $|z|=3$.

I see that the function has 2 singularities, at 0 and 2, so I need to find the residue of each. By examining the Laurent series, I have the following:
$$f(z)=\left(\frac{z+1}{z}\right)\left(\frac{1}{z-2}\right)=\left(\frac{z+1}{z^2}\right)\left(\frac{1}{1-2/z}\right)=\left(\frac{1}{z}+\frac{1}{z^2}\right)\left(\frac{1}{1-2/z}\right)$$ and therefore $$f(z)=\left(\frac{1}{z}+\frac{1}{z^2}\right)\left(1-\frac{2}{z}+\frac{4}{z^2}-\frac{8}{z^3}+\cdots\right)=\frac{1}{z}-\frac{1}{z^2}+\frac{2}{z^3}-\frac{4}{z^4}+\cdots$$
so the residue at 0 is 1.
Similarly, $$f(z)=\left(\frac{z+1}{2(z-2)}\right)\left(\frac{1}{1+(z-2)/2}\right)=\left(\frac{1}{2}+\frac{3}{2(z-2)}\right)\left(\frac{1}{1+(z-2)/2}\right)$$ and so $$f(z)=\left(\frac{1}{2}+\frac{3}{2(z-2)}\right)\left(1-\frac{z-2}{2}+\frac{(z-2)^2}{4}-\frac{(z-2)^3}{8}+\cdots\right)$$
Thus $$f(z)=\frac{3}{2(z-2)}-\frac{1}{4}+\frac{1}{8}(z-2)-\frac{1}{16}(z-2)^2+\cdots$$
and so the residue at 2 is $\frac{3}{2}$.
So I think $\int_Cf(z)dz=2\pi i(1+\frac{3}{2})= 5\pi i$, but that's not what the book is telling me - the book says the answer should be $2\pi i$. What am I doing wrong?
 A: Everything about your approach is fine, and your method of finding the Laurent series by factoring out a geometric series is smart.  You just made a mistake when you equated
$$\frac{1}{1-\frac{2}{z}} = 1-\frac{2}{z}+\frac{4}{z^2}-\frac{8}{z^3}+\cdots$$
The series on the right is actually an expansion of $\frac{1}{1+\frac{2}{z}}$, and it's the expansion at $z = \infty$ whereas you need the expansion at $z = 0$.
So what you should have is:
$$\frac{1}{1-\frac{2}{z}} = - \frac{z}{2} - \frac{z^2}{4} - \ldots -  \frac{z^n}{2^n} + \ldots$$
Then you'll get the Laurent series expansion at $z = 0$,
$$\frac{z+1}{z^2 - 2z} = -\frac{1}{2z} - \frac{3}{4} - \frac{3 z}{8} - \frac{3 z^2}{16} - \ldots - \frac{3z^n}{2^{n+2}} - \ldots$$ and the residue
you find this way is correct.
A: You can calculate the partial fraction decomposition to avoid calculating the Laurent's series :
$$\int_C\frac{z+1}{z^2-2z}dz=\frac32\int_C\frac{1}{z-2}dz-\frac12\int_C\frac{1}{z}dz$$
Then we apply Cauchy formula :
$$\int_C\frac{1}{z-2}dz=2\pi i=\int_C\frac{1}{z}dz$$
Or with Cauchy residue's theorem, $\frac{1}{z-2}$ is holomorphic inside $C$ except in $z=2$, so its Laurent series around $2$ is $\frac{1}{z-2}+\sum_{i=0}^\infty0z^i=\frac{1}{z-2}$. So the residue of $f$ around $2$ is $1$, so $\int_C\frac{1}{z-2}dz=2\pi i$. You can do the same with other integral.
So :
$$\int_C\frac{z+1}{z^2-2z}dz=2\pi i$$
A: You can do this without calculating any residues. For any $R>2,$ Cauchy's theorem shows your integral equals
$$\int_{|z|=R} \frac{z+1}{z^2-2z}\, dz = \int_0^{2\pi} \frac{(Re^{it} + 1)iRe^{it}}{R^2e^{2it} - 2Re^{it}}\, dt.$$
Do a little work to see this equals
$$\tag 1  i\int_0^{2\pi} \frac{1 + e^{-it}/R}{1 - 2e^{-it}/R}\, dt.$$
Remember, this stays the same for any $R>2.$ The integrands in $(1)$ $\to 1$ uniformly on $[0,2\pi]$  as $R\to \infty.$ Hence $(1)\to i\cdot 2\pi = 2\pi i$ as $R\to \infty.$ It follows that $2\pi i$ is the value of your integral.
