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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!

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  • $\begingroup$ Minor note: There need not be a curve connecting $0$ and $\infty$ in $(\mathbb{C}\cup \{\infty\})\setminus U$. That set is connected, but not necessarily path-connected. Consider the topologist's sine curve for a simple example. $\endgroup$ – Daniel Fischer Jan 31 '17 at 12:49
  • $\begingroup$ Thanks! So it turns out my lecturer used path-connect-ness as opposed to connectness in the defintion of a domain! Didn't know this difference in definition until I decided to look it up on the internet when I saw your comment. $\endgroup$ – Equivalent Triangle Jan 31 '17 at 13:30
  • $\begingroup$ An open subset of the plane is connected if and only if it is path-connected. So it doesn't matter whether we define a domain as a connected or a path-connected open set, both definitions are equivalent. It's the complement, $F = (\mathbb{C}\cup \{\infty\}) \setminus U$ which can be connected but not path-connected. Unless $U = \varnothing$, $F$ is not an open subset of the sphere, and it's only for open subsets that connectedness and path-connectedness are equivalent. The complement (in the sphere) of a simply connected domain is connected, but not necessarily path-connected. $\endgroup$ – Daniel Fischer Jan 31 '17 at 13:54
  • $\begingroup$ I see, thank you very much! $\endgroup$ – Equivalent Triangle Jan 31 '17 at 15:55
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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 $1$.

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  • $\begingroup$ With the Ln you defined (with constant adjusted), how can one show that $e^{\text{Ln}(w)}=w$ for all $w\in U$? $\endgroup$ – Equivalent Triangle Jan 31 '17 at 14:31
  • $\begingroup$ Sure, I have added that part too, there is a caveat in the first part though, for it to make sense you need $w$ and $e^w$ to both lie in $U$. Then in some open set around $w$ you find that $\mathrm{Ln}(e^w)$ differs from $w$ by a constant. $\endgroup$ – s.harp Jan 31 '17 at 14:42
  • $\begingroup$ Oops, I made some misjudgements. To see the relation you don't need both $w$ and $e^w$ to lie in $U$, my bad. I have corrected the answer. $\endgroup$ – s.harp Jan 31 '17 at 14:50
  • $\begingroup$ Thank you very much! btw it seems like the argument would be easier if at the start we define $\text{Ln}(w):=q+\int_{\gamma(w)}1/z\;dz$ where $q$ is a fix complex number such that $e^q=p$. That way we don't have to attempt to fix the constant of Ln in every path-component of $V=\{z\in\mathbb C: e^z\in U\}$. $\endgroup$ – Equivalent Triangle Jan 31 '17 at 15:54
  • $\begingroup$ $V$ is connected (if $V=V_1\cup V_2$ is the disjoint union of two open sets it follows $U=\exp(V_1)\cup\exp(V_2)$ is the disjoint union of two open sets since holomorphic maps are open maps). $\endgroup$ – s.harp Jan 31 '17 at 16:23

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