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Let $\gamma: [a,b] \rightarrow \mathbb{C}$ be a closed curve with $0 \notin \gamma([a,b]).$ Compute the Cauchy integral of $f(z):=\frac{1}{z}$ with regards to $\gamma$.

I have to compute

$$\begin{align} F(z)=\frac{1}{2 \pi i} \int _{\gamma}\frac{f(w)}{w-z}dw=\frac{1}{2 \pi i} \int _{\gamma}\frac{1}{w^2-wz}dw \end{align}$$

but as the curve is arbitrary I don't know how to go on. Can you guys help me?

Kekks
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  • Why do you say you have to compute that integral? It doesn't look correct...And then: distinguish between closed curves enclosing zero and those that don't. And none going on zero, otherwise things go awry. – DonAntonio May 25 '19 at 21:45
  • In the lecture we defined the Cauchy integral of a function $f$ as $F(z)=\frac{1}{2 \pi i} \int _{\gamma}\frac{f(w)}{w-z}dw$. And how do I distinguish between these two types of curves? Don't I need a parametrization of them? – Kekks May 25 '19 at 21:51
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    I think you're mixing stuff here, or else I'm missing something. What you say is an important Cauchy Theorem for analytic functions on a curve and its interior: to find the function's values by means of integral. Yet you asked for the integral of a specific function: very different ! – DonAntonio May 25 '19 at 21:59
  • I just looked it up in the script again. We just defined $F(z)$ to be the Cauchy integral. – Kekks May 25 '19 at 22:18
  • Presumably we want $\gamma$ to be a piecewise smooth curve (or at least a rectifiable curve), and the integral will need to be phrased in terms of the winding number of $\gamma$ around $0$. – Ted Shifrin May 25 '19 at 23:52

2 Answers2

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We assume that $z \not \in \gamma ([a,b])$, otherwise the integral is not defined. We can then calculate $\frac{1}{2\pi i} \int_{\gamma} \frac{1}{w} \frac{1}{w-z} \, dz$ by the residue formula:

  • If $z = 0$ then $\frac{1}{w} \frac{1}{w-z} = \frac{1}{w^2}$ has residue zero.
  • If $z \neq 0$ and both $z$ and $0$ are inside the interior of $\gamma$, then $\frac{1}{w}\frac{1}{w-z}$ has two simple poles, with residue $\frac{1}{0-z}$ for the pole at 0 and $\frac{1}{z}$ for the pole at $z$, thus in total we get residue zero. Thus if only one of $0$ or $z$ is in the interior of $\gamma$, then we get respectively the residues $\frac{1}{-z}$ and $\frac{1}{z}$. If both $0$ and $z$ are outside of $\gamma$ then the integral is zero.
user685060
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If, as you say, you're not supposed to use the Cauchy theorem, then I guess you're supposed to use the parametrization of the curve.

For a given $\gamma: [a,b] \ni t \mapsto \gamma(t) \in \mathbb C$: $$ F_\gamma(z) = \frac{1}{2\pi i}\int_\gamma \frac{1}{w(w-z)} dw = \frac{1}{2\pi i}\int_a^b \frac{1}{\gamma(t) \big(\gamma(t)-z\big)}\frac{d\gamma(t)}{dt}dt$$ You cannot actualy calculate this integral unless you knwo what the curve is, but given a parametrization, it can be done for any curve, closed or not.

Adam Latosiński
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