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 \(\Omega?\)

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 \[ \phi(z) = \frac{1}{i2\pi} \int_{\gamma} -\frac{1/z^{|n|}}{w^{-1}} dw + \frac{1}{i2\pi} \int_{\gamma} \frac{1/z^{|n|}}{w-z} dw = 0. \]