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Many functions can be analytically continued to $\mathbb C$ except the branch cut.

However, it appears to me that for every function $f(z)$ that has a branch cut, there always exists a non-constant meromorphic/entire function $g(z)$ such that $f(g(z))$ can be analytically continued to the whole $\mathbb C$.

For example, $$\sqrt {x^2}=x$$ $$\ln e^x=x$$ $$\arccos\cos x =x$$ $$\ln\ln e^{e^x}=x$$ $$\operatorname{W}(xe^x)=x$$ More complicated examples: $$f(x)=\sqrt{(x+1)(x+3)}=\sqrt{(x+2)^2-1}$$ $$g(x)=-2+\cosh x$$ $$f(x)=x^\alpha\qquad{\alpha\in\mathbb C}$$ $$g(x)=e^x$$

Is this true?

Szeto
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There is indeed always a function such as you describe.

First, pick an open disk in $\mathbb{C}$ where $f$ is analytic. Since $f$ is non-constant, there is a point $a$ in this region where $f'(a)\neq 0$. Thus, by the Lagrange Inversion Theorem, there is a local inverse $g(z)$ which is analytic in a neighborhood of $f(a)$. In this neighborhood, $f(g(z))=z$, and the identity function $z$ can of course be extended to the complex plane.

Now, as you likely noticed, this restriction of the domain to an open disk before analytic continuation is necessary. Any $g(z)$ such that $f(g(z))$ has no branch cut must avoid the cut of $f$. By Picard's theorem there is no entire function on $\mathbb{C}$ which can avoid more than one point, and no meromorphic function on $\mathbb{C}$ which can avoid more than two. Thus for any entire or meromorphic $g$, either $g$ is constant or the preimage of the branch cut under $g$ must be a branch cut for $g$. The second case contradicts the meromorphicity of $g$. Thus there is no non-constant meromorphic $g$ such that $f(g(z))$ is meromorphic on all of $\mathbb{C}$.

Grant B.
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  • Thank you. I will think about it for a while to see if I have any doubts. – Szeto Aug 01 '18 at 08:15
  • ‘Any $g(z)$ such that $f(g(z))$ has no branch cut must avoid the cut of $f$.’ Umm...not necessarily. In my first example, $x^2$ does not avoid $0$, which is the branch point of $\sqrt x$. – Szeto Aug 01 '18 at 09:50
  • Using the standard branch cut for $\sqrt{}$ along the negative real axis, the identity $\sqrt{x^2}=x$ is only valid for $\text{Re}(x)>0$. (And the point $x=0$ as you said, but no neighborhood around it.) The discontinuity in the composite function is along the line $\text{Re}(x)=0$. – Grant B. Aug 01 '18 at 10:22
  • To clarify, by "avoid the cut", I mean there must not exist a path in $\mathbb{C}$ such that the image of the path under $g$ crosses the branch cut. For $x^2$ any path crossing the imaginary axis will cross the branch cut of $\sqrt{}$, so $\sqrt{x^2}$ is discontinuous along the imaginary axis. – Grant B. Aug 01 '18 at 20:12
  • By the way, is it guaranteed that $g’(z)$ can be analytically continued to the whole complex plane? – Szeto Aug 01 '18 at 23:32
  • If $g(z)$ can, then so can all derivatives of $g$. Generically a local inverse will not be capable of extending to the entire plane (though there are plenty of exceptions), but if there are a finite number of branch cuts we can remove them via coordinate transforms that bring each point to the origin and rotate the cut to be along the negative real axis, then letting $z = e^{z'}$. This removes one branch cut at a time while keeping $g$ entire (though as described the composition $f(g(z))$ will only be analytic in a subset of $\mathbb{C}$). – Grant B. Aug 02 '18 at 03:41
  • Sorry, I mean ‘is it guaranteed that $g’(z)$ can be meromorphically continued to the whole complex plane’. However, as how I interpret your response, it is likely that a branch cut of $g$ is inevitable. – Szeto Aug 02 '18 at 04:24