$$\frac{1}{2z-1} = f(z)$$
So I don't think CR theoreum applies. So if the denominator is 0, then it's not defined. If z = 1/2, then it is not defined at 0. So f(z) it is not analytic even within C right? So there does not exist a simply connected domain D on which f is analytic and such that D contains C. So cannot apply Cauchy's Riemann equation right?
So we have to integrate another way.
So unit circle could be expressed as $e^{it} = z(t)$ where t is between 0 and $2\pi$
Furthermore $dz/dt = ie^{it}$
$$\int_0^{2\pi} \frac{1}{2e^{it} - 1} \cdot ie^{it}$$
Am I on the right track? If so, can someone help with the integral?
METHODOLOGY $1$: CONTOUR DEFORMATION
We can deform the contour $|z|=1$ and apply Cauchy's Integral Theorem to assert that for $0<r<1/2$
$$\begin{align} \oint_{|z|=1}\frac1{2z-1}\,dz&=\oint_{|z-1/2|=r}\frac1{2z-1}\,dz\\\\ &=\int_0^{2\pi}\frac1{2re^{it}}ire^{it}\,dt\\\\ &=i\pi \end{align}$$
METHODOLOGY $2$: PARAMETERIZATION
Alternatively, we can parameterize the contour $|z|=1$ and write
$$\begin{align} \oint_{|z|=1}\frac1{2z-1}\,dz&=\lim_{\varepsilon\to 0^+}\int_{\varepsilon}^{2\pi-\varepsilon}\frac{ie^{it}}{2e^{it}-1}\,dt\\\\ &=\lim_{\varepsilon\to 0^+}\int_{\varepsilon}^{2\pi-\varepsilon}\frac{\sin(t)+i2(1-\cos(t))}{5-4\cos(t)}\,dt\\\\ &=\lim_{\varepsilon\to 0^+}\left.\left(\frac14\log(5-4\cos(t))+i2\left(\frac t4-\frac16 \arctan\left(3\tan(t/2)\right)\right)\right) \right|_{\varepsilon}^{2\pi -\varepsilon}\\\\ &=i\pi \end{align}$$
as expected!
METHODOLOGY $3$: COMPLEX LOGARITHM
As a third approach, we cut the plane so that the natural logarithm is defined as $\log(z)=\text{Log}(|z|)+i\arg(z)$, where $0\le \arg(z)<2\pi$ we have
$$\oint_{|z|=1}\frac1{2z-1}\,dz=\lim_{\varepsilon\to 0^+}\left.\left(\frac12\log(2z-1)\right)\right|_{z=1+i\varepsilon}^{z=1-i\varepsilon}=i\pi$$