If \(n=-1\), no primitive exist: the function \(\gamma: t \in [0,1] \to e^{i2\pi t}\) is a closed rectifiable path of \(\mathbb{C}^*\) and \[ \int_{\gamma} \frac{dz}{z} = \int_0^1 \frac{e^{i2\pi t} i2\pi}{e^{i2 \pi t}} \, dt = i2\pi, \] which is nonzero.

The function \[ g(z) = \log \frac{z-1}{z} = \log \left(1 - \frac{1}{z} \right) \] is defined on \(\mathbb{C} \setminus [0,1]\) and is a primitive of \(f\). Indeed \(g(z)\) is defined as long as neither of the conditions \(z = 0\) and \(1 - 1/z \in \mathbb{R}_-\) are met; they are equivalent to the condition \(z \in [0,1]\), which is excluded. Moreover, \(g\) satisfies \[ g'(z) = \frac{1/z^2}{1 - 1/z} = \frac{1}{z(z-1)} \] hence it is a primitive of \(f\).

1. Show that \(\beta = \alpha \circ \phi\) is a rectifiable path which has the same initial point, terminal point and image as \(\alpha\).

2. Prove that for any continuous function \(f: \alpha([0,1]) \to \mathbb{C}\), \[ \int_{\alpha} f(z) \, dz = \int_{\beta} f(z) \, dz. \]

3. Prove that the paths \(\alpha\) and \(\beta\) have the same length.

Answers

1. The statement about the initial and terminal points is obvious. The one relative to the image holds because, under the assumptions that were made, the function \(\phi\) is a bijection from \([0,1]\) on itself (and its inverse is also continuously differentiable).

2. We have \[ \int_{\beta} f(z) \, dz = \int_{0}^1 (f \circ \beta)(t) \beta'(t)\,dt = \int_{0}^1 (f \circ \alpha)(\phi(t)) \alpha'(\phi(t))\,(\phi'(t) dt). \] The change of variable \(s = \phi(t)\) leads to \[ \int_{\beta} f(z) \, dz = \int_{0}^1 (f \circ \alpha)(s) \alpha'(s)\,ds = \int_{\alpha} f(z) \, dz. \]

3. We have \[ \int_0^1 |\beta'(t)|\, dt = \int_{0}^1 |\alpha' (\phi(t)) \phi'(t)| \, dt = \int_{0}^1 |\alpha' (\phi(t))| \, \phi'(t) dt \] The change of variable \(s = \phi(t)\) leads to \[ \int_0^1 |\beta'(t)|\, dt = \int_0^1 |\alpha'(s)| \, ds, \] hence the lengths of \(\alpha\) and \(\beta\) are equal.

Alternatively, we can build explicitly such a primitive: the function \[ f: z \mapsto \log (z e^{i(\pi - \alpha)}); \] it is defined and holomorphic on \(\mathbb{C}_{\alpha}\) and for any \(z \in \mathbb{C}_{\alpha}\), \[ f'(z) = \frac{1}{z e^{i(\pi - \alpha)}} \times e^{i(\pi - \alpha)}= \frac{1}{z}. \]