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?
Best Answer
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.