Define a continuous log in a simply connected domain not equal to the whole of $\mathbb C$

Question

So I know that in the complex plane given any straight line $L$, say from the origin to infinity, one can define a continuous log on $\mathbb C\setminus L$. It seems to me that if we have a simply connected domain $U\subset\mathbb C$ such that $0\not\in U$, then we can define a continuous log on $U$. Because by simply-connect-ness we can find a curve staring from 0 that goes to infinity, which is not in $U$, so we have a "slit" like before, hence we should be able to define a continuous log. However I find it very hard to write down a rigorous proof. So I want to know, is this true? If it is, how can I write down a proof? Thanks!

Answer 1

Let $U$ be a simply connected open domain that does not contain $0$. Then the map $$z\mapsto \frac1z$$ is holomorphic on $U$ and for any two differentiable paths $\gamma_1,\gamma_2$ in $U$ whose end-points agree you have: $$\int_{\gamma_1}\frac1z\, dz=\int_{\gamma_2}\frac1z\,dz$$ This follows from $U$ being simply connected, ie every loop in $U$ is contractible. Pick a starting point $p$ and for any point $w\in U$ let $\gamma(w)$ be a differentiable path from $p$ to $w$, it follows that: $$\mathrm{Ln}(w):=\int_{\gamma(w)}\frac1z\,dz$$ is well defined. It is also a holomorphic function, where the existence of derivative can be shown by taking $\gamma(w+h)$ to be $\gamma(w)$ followed by a straight line from $w$ to $h$ and using the usual argument to find that the limit $\lim_{h\to0}\frac{\mathrm{Ln}(w+h)-\mathrm{Ln}(w)}h=\frac1w$.

That relation shows that this function does act as a logarithm in the sense that: $$\frac{d}{dw}(\mathrm{Ln}(e^{w})-w)=\frac{e^w}{e^w}-1=0$$ and $\mathrm{Ln}(e^w)$ differes from $w$ only by a constant. (This assumes that $e^w$ lies in $U$.) One also has: $$\frac d{dw}\frac{e^{\mathrm{Ln}(w)}}w=\frac{e^{\mathrm{Ln}(w)}}{w^2}-\frac{e^{\mathrm{Ln}(w)}}{w^2}=0$$ So $e^{\mathrm{Ln}(w)}$ must be a constant times $w$ (this requires $w$ to lie in $U$). You can combine this conclusion with the one before to find a relation between the additive and the multiplicative constant for such $w$ so that $e^w$ lies un $U$: $$Me^w\overset!=\exp({\mathrm{Ln}(e^w)})\overset!=e^{w+A}$$ If you have chosen the additive constant to be $0$ or a factor of $2\pi i$ then the multpilicative constant must be a $1$.