# Exercises

## Cauchy’s Converse Integral Theorem

### Question

Let $$\Omega$$ be an open subset of $$\mathbb{C}$$.

Suppose that for every holomorphic function $$f: \Omega \to \mathbb{C},$$ $\int_{\gamma} f(z) \, dz = 0$ for some sequence $$\gamma$$ of closed rectifiable paths of $$\Omega.$$ What conclusion can we draw? What if the property holds for every sequence $$\gamma$$ of closed rectifiable paths of $$\Omega?$$

For any $$w \in \mathbb{C} \setminus \Omega$$, the function $$f: z \in \Omega \mapsto 1/(z-w)$$ is defined and holomorphic, thus $\mathrm{ind}(\gamma, w) = \frac{1}{i2\pi} \int_{\gamma} \frac{dz}{z-w} = 0$ and therefore $$\mathrm{Int} \, \gamma \subset \Omega$$. Now, suppose that this conclusion holds for any sequence $$\gamma$$ of closed rectifiable paths of $$\Omega.$$ Since the winding number is locally constant and since for any closed path $$\mu$$ of $$\Omega$$ and any $$\epsilon > 0$$, there is a closed rectifiable path $$\gamma$$ of $$\Omega$$ such that $\forall \, t \in [0,1], \; |\gamma(t) - \mu(t)| < \epsilon,$ we also have $$\mathrm{ind}(\mu, w) = \mathrm{ind}(\gamma, w) = 0$$. Therefore $$\mathrm{Int} \, \mu \subset \Omega$$: the set $$\Omega$$ is simply connected.

## Cauchy Transform of Power Functions

### Question

Compute for any $$n \in \mathbb{Z}$$ and any $$z \in \mathbb{C}$$ such that $$|z| \neq 1$$ the line integral $\phi(z) = \frac{1}{i2\pi} \int_{[\circlearrowleft]} \frac{w^n}{w - z} dw.$

For any $$z \in \mathbb{C}$$ such that $$|z| \neq 1$$, the function $\psi_z: w \mapsto \frac{w^n}{w - z}$ is defined and holomorphic on $$\Omega = \mathbb{C} \setminus \{z\}$$ if $$n \geq 0$$; it is defined and holomorphic on $$\Omega = \mathbb{C} \setminus \{0, z\}$$ if $$n < 0$$. The interior of $$[\circlearrowleft]$$ is the open unit disk.

We now study separately four configurations.

1. Assume that $$n \geq 0$$ and $$|z| > 1$$. The interior of $$[\circlearrowleft]$$ is included in $$\Omega$$, hence by Cauchy’s integral theorem, $$\phi(z) = 0$$.

Alternatively, Cauchy’s formula was also applicable.

2. Assume that $$n \geq 0$$ and $$|z| < 1$$. The unique singularity of $$\psi_z$$ is $$w=z$$; it satisfies $$\mathrm{ind} ([\circlearrowleft], z) = 1$$. Let $$\gamma(r) = z + r [\circlearrowleft]$$; we have $\mathrm{res}(\psi_z, z) = \lim_{r \to 0} \frac{1}{i 2 \pi} \int_{\gamma(r)} \frac{w^n}{w - z} dw = \lim_{r \to 0} \int_0^1 (z + r e^{i2\pi t})^n dt = z^n,$ hence by the residues theorem, $$\phi(z) = z^n$$.

Alternatively, Cauchy’s formula was also applicable.

3. Assume that $$n < 0$$ and $$|z| > 1$$. We have $\phi(z) = \frac{1}{i2\pi} \int_{[\circlearrowleft]} \frac{1}{w^{|n|}(w-z)} \, dw.$ If $$n=-1$$, Cauchy’s formula provides the answer: $\phi(z) = \frac{1}{0-z} = - z^{-1}.$ Otherwise $$n < -1$$, we may use integration by parts (several times): $\begin{split} \phi(z) = &\, \frac{1}{i 2 \pi} \int_{[\circlearrowleft]} \frac{1}{w^{|n|}(w-z)} \, dw \\ = &\, - \frac{1}{i 2 \pi} \int_{[\circlearrowleft]} \frac{-1}{|n| - 1} \frac{1}{w^{|n|-1}} \frac{-1}{(w - z)^2}\, dw \\ = &\, \dots \\ = &\, (-1)^{|n|-1}\frac{1}{i 2 \pi} \int_{[\circlearrowleft]} \frac{1}{(|n| - 1)!} \frac{1}{w} \frac{(|n|-1)!}{(w - z)^{|n|}} \, dw \\ = &\, - \frac{1}{i 2 \pi} \int_{[\circlearrowleft]} \frac{1}{w} \frac{1}{(z - w)^{|n|}} \, dw. \end{split}$ At this point, Cauchy’s formula may be used again and we obtain $\phi(z) = - z^{n}.$

Alternatively, we may perform the change of variable $$w = 1/\xi:$$ $\begin{split} \phi(z) = &\, \frac{1}{i2\pi} \int_{[\circlearrowleft]} \frac{w^n}{w - z} dw \\ = &\, - \frac{1}{i2\pi} \int_{[\circlearrowleft]} \frac{\xi^{-n} }{\xi^{-1} - z} \left(-\frac{d\xi}{\xi^2}\right) \\ = &\, - \frac{1}{z} \frac{1}{i2\pi} \int_{[\circlearrowleft]} \frac{\xi^{-n-1}}{\xi - z^{-1}}d\xi. \end{split}$ As $$-n-1 \geq 0$$ and $$|z^{-1}| < 1$$, we may invoke the result obtained for the first configuration: it provides $$\phi(z) = -z^n$$.

Alternatively, we may perform a partial fraction decomposition of $$w \mapsto 1/(w^{|n|}(w-z))$$. Since $1 - \left( \frac{w}{z} \right)^{|n|} = \left(1 - \frac{w}{z} \right) \left(1 + \frac{w}{z} + \dots + \left(\frac{w}{z}\right)^{|n| -1} \right),$ we have $\frac{1}{w-z} = -\frac{1}{z} \left(1 + \frac{w}{z} + \dots + \left(\frac{w}{z}\right)^{|n| -1} \right) + \frac{w^{|n|}/z^{|n|}}{w-z}$ and therefore $\frac{1}{w^{|n|}(w-z)} = - \left( \frac{z}{w^{|n|}} + \frac{1/z^2}{w^{|n|-1}} + \dots + \frac{1/z^{|n|}}{w^{-1}} \right) + \frac{1/z^{|n|}}{w-z}.$ The integral along $$\gamma$$ of $$w \in \mathbb{C} \mapsto 1/w^p$$ is zero for $$p>1$$ since this function has a primitive. The integral of $$w \mapsto 1/(w-z)$$ is also zero since $$|z|>1$$. Finally, $\phi(z) = \frac{1}{i2\pi} \int_{\gamma} -\frac{1/z^{|n|}}{w^{-1}} dw = -z^n.$

4. Assume that $$n < 0$$ and $$|z| < 1$$. There are two singularities of $$\psi_z$$ in the interior of $$[\circlearrowleft]$$, $$w=0$$ and $$w=z$$, unless of course if $$z=0$$.

If $$z=0$$, we have $\phi(z) = \frac{1}{i2\pi} \int_{[\circlearrowleft]} w^{n-1} dw = 0$ because $$w \in \mathbb{C}^* \mapsto w^{n}/n$$ is a primitive of $$w \in \mathbb{C}^* \mapsto w^{n-1}$$.

We now assume that $$z \neq 0$$. The residue associated to $$w=z$$ can be computed directly with Cauchy’s formula; with $$\gamma(r) = z + r[\circlearrowleft]$$, we have $\mathrm{res}(\psi_z, z) = \lim_{r \to 0} \frac{1}{i2\pi} \int_{\gamma(r)} \frac{w^n}{(w-z)} dw = z^n.$ On the other hand, using computations similar to those of the previous question, we can derive $\mathrm{res}(\psi_z, 0) = \lim_{r \to 0} \frac{1}{i2\pi} \int_{r[\circlearrowleft]} \frac{w^n}{(w-z)} dw = - z^n.$ Consequently, $$\phi(z) = 0$$.

In the case $$z \neq 0$$, we may perform again the change of variable $$w = 1/\xi$$ that provides $\phi(z)= - \frac{1}{z} \frac{1}{i2\pi} \int_{[\circlearrowleft]} \frac{\xi^{-n-1}}{\xi - z^{-1}}d\xi.$ As $$-n-1 \geq 0$$ and $$|z^{-1}| > 1$$, we may invoke the result obtained for the first configuration: it yields $$\phi(z) = 0$$.

There is yet another method: we can notice that for $$r>1$$, the interior of the path sequence $$(r [\circlearrowleft], [\circlearrowleft]^{\leftarrow})$$, which is the annulus $$\{z \in \mathbb{C} \; | \; 1 < |z| < r\},$$ is included in $$\Omega$$. Cauchy’s integral theorem provides $\forall \, r > 1, \; \phi(z) = \frac{1}{i2\pi} \int_{r[\circlearrowleft]} \frac{w^n}{w - z} dw.$ and the M-L estimation lemma $\forall \, r > 1, \; |\phi(z)| \leq \frac{1}{r^{|n|-1}(r - |z|)}.$ The limit of the right-hand side when $$r \to +\infty$$ yields $$\phi(z) = 0$$.

Finally, we may use again the partial fraction decomposition of $$w \mapsto 1/(w^{|n|}(w-z))$$: $\frac{1}{w^{|n|}(w-z)} = - \left( \frac{z}{w^{|n|}} + \frac{1/z^2}{w^{|n|-1}} + \dots + \frac{1/z^{|n|}}{w^{-1}} \right) + \frac{1/z^{|n|}}{w-z}.$ The integral along $$\gamma$$ of $$w \in \mathbb{C} \mapsto 1/w^p$$ is zero for $$p>1$$ since this function has a primitive. Therefore $\phi(z) = \frac{1}{i2\pi} \int_{\gamma} -\frac{1/z^{|n|}}{w^{-1}} dw + \frac{1}{i2\pi} \int_{\gamma} \frac{1/z^{|n|}}{w-z} dw = 0.$