I have a question about a step in the proof of the following theorem.
Theorem: $Ln: \mathbb{C}- \{x \in \mathbb{R}: x \leq 0\} \to \mathbb{C}: z = re^{i \phi} \mapsto \ln r + i \phi$ $(\phi \in (-\pi, \pi), r > 0)$ is analytic.
Proof: We have
$$\begin{cases} u = \ln r = 1/2\ln (x^2 + y^2) \\ v = \phi\end{cases}$$
and it follows immediately that
$$\partial{u}/\partial x = x/(x^2 + y ^2), \partial{u}/\partial x = y/(x^2 + y ^2)$$
$$\begin{cases} u = \ln r = 1/2\ln (x^2 + y^2) \\ v = \phi\end{cases}$$ We have
$$\begin{cases} x = r \cos \phi \\ y = r \sin \phi\end{cases}$$
Differentiating for $x$:
$$\begin{cases} 1 = \cos \phi \partial r / \partial x - r \sin \phi \partial \phi/ \partial x \\ 0 = \sin \phi \partial r / \partial x + r \cos \phi \partial \phi/ \partial x\end{cases}$$
and solving this:
$$\partial \phi/\partial x = - \sin \phi/r = -y/(x^2+y^2)$$
and the Cauchy-Riemann conditions are satisfied.
Questions:
Why can we take the derivative of both sides? Don't we use implicit differentiation there? I.e., do we have to use the theorem of the implicit functions to make this proof work?
