Does the branch of the initial value determine the branch for the whole curve?

Question

Let $\Omega$ be a connected and simply-connected open subset of $\mathbf{C}$ which does not contain the origin. Let $\gamma :[0,1] \to \Omega$ be a simple smooth curve and let $\zeta := \gamma(0)$. Let $\phi:[0,1] \to \mathbf{C}$ be a smooth map such that $e^{\phi(t)} = \gamma(t)$ for all $t \in [0,1]$ (i.e., at each point $\phi$ is a logarithm of $\gamma$). Since $\Omega$ is connected and simply-connected, there exists a unique single-valued branch $"f"$ of the logarithm defined on $\Omega$ such that $f(\zeta) = \phi(0)$. Does it necessarily follow that $\phi$ is the restriction of $f$ to the curve $\gamma$?

Answer 1

Hint

The answer is positive and follows from:

• The fact that a branch of the logarithm is uniquely defined on an open disk $D(a,r)$ such that $0 \notin D(a,r)$ by its value $f(a)$ at $a$.
• The image of $[0,1]$ under $\gamma$ is path connected, hence connected.
• Moreover this image is included in the simply connected open subset $\Omega$ for which $0 \notin \Omega$.

Answer 2

Yes. For all $t \in [0, 1]$ is $$ e^{\phi(t) - f(\gamma(t))} = \frac{e^{\phi(t)}}{e^{f(\gamma(t))}} = \frac{\gamma(t)}{\gamma(t)} = 1 $$ because $e^{f(w)} = w$ holds for all $w \in \Omega$. It follows that the continuous function $$ [0, 1] \ni t \mapsto \frac{1}{2 \pi i }(\phi(t) - f(\gamma(t))) $$ takes only integer values on the connected interval $[0, 1]$, and is therefore constant. But $\phi(0) = f(\gamma(0))$ per the construction, so that $\phi(t) = f(\gamma(t))$ holds for all $t$.