Contour integral of $z\mapsto \frac{1}{z(z-1)}$ over the positively oriented circle with radius $\frac12$

Question

I have tried the residue theorem since the only problem lies at $0$ since $1$ is outside the circle . I get $f(z)=\frac{1}{z-1}$ with $f(0)=-1$ and so I get that the integral is equal to $-2πi$. Something feels wrong since he suggests splitting it in to $\frac1{z-1}-\frac1{z}$. I dont get why the residue theorem wouldnt apply though.Any answer is much appreciated.

Answer 1

There is nothing wrong with your approach. However, if you split your fraction like that, you get the correct answer in a more elementary way. Using that approach, you get (assuming that $\gamma(t)=\frac12e^{it}$, with $t\in[0,2\pi]$)\begin{align}\int_\gamma\frac1{z(z-1)}\,\mathrm dz&=\int_\gamma\frac1{z-1}\,\mathrm dz-\int_\gamma\frac1z\,\mathrm dz\\&=2\pi i\left(\operatorname{wind}(\gamma,1)-\operatorname{wind}(\gamma,0)\right)\\&=2\pi i(0-1)\\&=-2\pi i.\end{align}So, the residue theorem is not needed at all. All you need to know is Cauchy's integral formula.