Analytic branch of the complex logarithm in the logarithmic spiral branch cut We define the domain (a connected open set) to be the complex plane minus the logarithmic spiral:

*

*$D = \mathbb{C} \setminus \{te^{it} : t \ge 0 \} $.

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I need to find a branch $ F(z) $ of the complex logarithm $ \log(z) $ in $ D $ which satsfies $ F(1) = 0 $.
Now, my problem is that we cannot use a standard branch which takes off a ray in some direction starting from the origin, so I don't even know how to start thinking of a branch, let alone proving it is analytic afterwards.
I'd like to get some help, thank you in advance.
 A: Indeed, any non self-intersection curve that leaves a connected domain can be used as a branch cut for the complex logarithm, there is nothing special in a spiral or a straight half-line.
The idea of the branch cut is to define the value of the argument function continuously such that the logarithm would be an holomorphic function in the domain defined by the branch cut.
A continuous argument function for this branch cut is given by
$$
\arg_D(z):=\max\{\theta \in[0,\infty ):ze^{-i\theta }-\theta \in[0,\infty )\}
$$
The idea above is to pick the unique value of $\theta $ such that $z$ belongs to the segment between the points of the logarithmic spiral $\theta e^{i\theta}$ and $(\theta +2\pi)e^{i\theta }$. Then a complex logarithm compatible with this branch cut is given by
$$
\log _D(z):=\log |z|+i\arg_D(z)
$$
Finally observe that if $z=1$ we have that
$$
e^{-i\theta }-\theta\in[0,\infty ) \,\land\, \theta \geqslant 0\iff \theta ^{-i\theta }\geqslant \theta \geqslant 0\iff 1\geqslant  \theta e^{i\theta }\geqslant 0
$$
and so
$$
\arg_D(1)= \max\{\theta \in[0,\infty ): \theta e^{i\theta }\in[0 , 1]\}=\max\{\theta \in \pi \mathbb{Z}: \theta e^{i\theta }\in[0 , 1]\}=0
$$
so the above defined complex logarithm $\log _D$ have also the property that $\log _D(1)=0$.∎
A: Define
$$\phi_\pm : \mathbb C \to \mathbb C, \phi(z) = z e^{\pm i \lvert z \rvert}.$$
These maps are continuous and it is easy to verify that $\phi_- \circ \phi_+ = id$ and $\phi_+ \circ \phi_- = id$. Thus $\phi_\pm$ are homeomorphisms which are inverse to each other.
We have
$$\phi_+([0,\infty)) = \{\phi_+(t) \mid t \in [0,\infty) \} = \{te^{it} \mid t \in [0,\infty) \}$$
and therefore
$$\phi_+(\mathbb C \setminus [0,\infty))  = D .$$
This shows that $\phi_+$ maps the sliced plane $P = \mathbb C \setminus [0,\infty)$ homeomorphically onto $D$. Hence $D$ is simply connected which is sufficient to conclude that there exists a branch of the logarithm $\ln_D$ on $D$ such that $\ln_D(1) = 0$. Simply take $\ln_D(z) = \int_\gamma \frac 1 \zeta d\zeta$, where $\gamma$ is any curve from $1$ to $z$.
Update:
The essential ingredient in the above argument is that each holomorphic $g : U \to \mathbb C$ on a simply connected domain $U$ has an antiderivative $G : U \to \mathbb C$. Explicitly we can take
$G(z) = \int_\gamma g(\zeta)d\zeta$ where $\gamma$ is any curve from a fixed $z_0 \in U$ to $z \in U$. This $G$ has the property $G(z_0) = 0$.
In the above situation this is applied to $g(z) =  \frac 1 z$ and $z_0 = 1$. But why is the resulting antiderivative $\ln_D : D \to \mathbb C$ a branch of the logarìthm?
Let us consider a domain $U \subset \mathbb C^* = \mathbb C \setminus \{0\}$. A holomorphic function  $\ell : U \to \mathbb C$ is called a branch of the logarithm on $U$ if $e^{\ell(z)} = z$ for all $z \in U$. The chain rule shows that $e^{\ell(z)} \ell'(z) = 1$ which gives $\ell'(z) = \frac 1 z$. Thus each branch of the logarithm on $U$ is an antiderivative of $\frac 1 z$.
Given a domain $U \subset \mathbb C^*$ it is not obvious (i) that $\frac 1 z$ has an antiderivative  on $U$, and (ii) if such an antiderivative exists, that it is a branch of the logarithm on $U$.
A sufficient condition for the existence is that $U$ is simply connected. Even in that case not every antiderivative $\ell$ will be a branch of the logarithm on $U$. In fact, fixing an antiderivative $\ell_0$, the collection of all antiderivatives is given by the functions $\ell_c : U \to \mathbb C, \ell_c(z) = \ell_0(z) + c$, with $c \in \mathbb C$. We then get
$$e^{\ell_c(z)} = e^{\ell_0(z)}e^c \tag{1}$$
which can certainly not be a branch of the logarithm for all $c$. A minimal requirement for $\ell_c$ to be a branch of the logarithm is that $e^{\ell_c(z_0)} = z_0$ for some $z_0 \in U$. This means
$$e^c = \frac{z_0}{e^{\ell_0(z_0)}} . \tag{2}$$
Such $c$ always exists because the exponential function $\exp(z) = e^z$  has image $\mathbb C^*$. Thus there always exists an antiderivative $\ell$ such that $e^{\ell(z_0)} = z_0$. But now we have
Lemma. Let $\ell : U \to \mathbb C$ be an antiderivative on $\frac 1 z$ on $U$ such that $e^{\ell(z_0)} = z_0$ for some $z_0  \in U$. Then $\ell$ is a branch of the logarithm on $U$.
Proof. Consider the function $\phi : U \to \mathbb C, \phi(z) = \frac{e^{\ell(z)}}{z}$. We have
$$\phi'(z) = \frac{e^{\ell(z)}\frac 1 z z - e^{\ell(z)}}{z^2} = 0 .$$
Thus $\phi$ is constant. This proves the Lemma because $\phi(z_0) = 1$.
