Evaluate $\oint_C\frac{dz}{z-2}$ around the circle $|z-2| = 4$

Question

I don't completely understand how to approach these questions. I suppose the notation $\oint_C$ is something I'm not used to.

So far, I have $\oint_C\frac{dz}{z-2} = \log(z-2)$. From here, I suppose I could utilize $|z-2| = 4$ and $z-2 = re^{i\theta}$ somehow, but I'm not completely sure.

Is it the matter of that that $r = 4$, and $\theta = 2\pi$ and $\theta = 0$ (considering we're going over the entire region? Then, we have $\log(4e^{i2\pi}) - \log(4e^{i0}) = 2\pi i+\log(4) - (i0 + \log(4)) = 2\pi i $

I think I may be right. I know the answer is correct, but I don't know if my approach was necessarily correct.

Answer 1

By definition,

$$\oint_C f(z) dz = \int_a^b f(\gamma(t)) \gamma'(t) dt$$

where $\gamma : [a, b] \to \mathbb{C}$ is a parameterization of $C$ (provided $C$ is nice, as occurs in this case). This is simply a matter of definition.

Now in order to compute the integral, finding an antiderivative won't really work as you've tried. The problem is that $\log{(z - 2)}$ cannot be made holomorphic (differentiable) away from $2$ - you'll still have to have a branch cut, so it's not even an antiderivative anymore. Note that $e^{2\pi i} = e^0$ in the complex plane, so if we have

$$\ln{e^{2\pi i}} = 2\pi i \text{ while } \ln{1} = 0$$ then $\ln$ isn't even a single-valued function anymore. This is why we'd have to take a branch cut or view the logarithm as multi-valued.

The general technique for solving these problems is to write down a parameterization and make it an integral over some real interval. A circle can be parameterized very nicely; let's take

$$\gamma : [0, 2\pi] \to\mathbb{C}$$

defined by $$\gamma(t) = 2 + 4e^{it}$$

Then

\begin{align*} \oint_C f(z) dz &= \int_0^{2\pi} \frac{1}{(2 + 4e^{it}) - 2} 4ie^{it} dt \\ &= \int_0^{2\pi} \frac{4ie^{it}}{4e^{it}} dt \\ &= i \int_0^{2\pi} dt = 2\pi i \end{align*}

Answer 2

You must know, I presume, the definition of line integral (as we need it for complex integrals)...right? Well, parametrize the given path as what it is: a circle of radius $\;2\;$ centered at $\;2\;$), and also

$$|z-2|=4\iff z=2+4e^{it}\;,\;\;t\in[0,2\pi]\;,\;\;dz=4ie^{it}\;,\;\;\text{so :}$$

$$\oint\limits_{|z-2|=4}\frac{dz}{z-2}=\int\limits_0^{2\pi}\frac{4ie^{it}}{4e^{it}}dt=2\pi i$$

Answer 3

Using Cauchy integral formula,

Here $f(z)=1,z_0=2$ is isolated singular point,$f(z)$ is analytic on and within $C:|z-2|=4$

$$\oint\limits_{|z-2|=4}\frac{dz}{z-2}=2\pi i f(z_0=2)=2\pi i$$

Answer 4

As one should know is that $\dfrac{1}{2\pi i}\int_c\dfrac{dz}{z-p}$ is the winding number of the closed curve $c$ around a point $p$. So the given integral equals $2n\pi i$, where $n$ is number of windings, depending on the parametrization of $c$.