1

$$\frac{1}{2z-1} = f(z)$$

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?

Jwan622
  • 5,704

2 Answers2

1

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

$$\oint_{|z|=1}\frac1{2z-1}\,dz=\lim_{\varepsilon\to 0^+}\left.\left(\frac12\log(2z-1)\right)\right|_{z=1+i\varepsilon}^{z=1-i\varepsilon}=i\pi$$

Mark Viola
  • 179,405
0

Using residue theorem we easily get $$\oint\limits_C {\frac{{dz}}{{2z - 1}} = 2\pi i}\;\text{Res}\left(\frac{1}{2 z-1},\left\{z,\frac{1}{2}\right\}\right) =\pi i$$

Because $$\text{Res}\left(\frac{1}{2 z-1},\left\{z,\frac{1}{2}\right\}\right)=\underset{z\to \frac{1}{2}}{\text{lim}}\frac{z-\frac{1}{2}}{2 z-1}=\frac{1}{2}$$

Raffaele
  • 26,371