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The book roughly states :-

Suppose an element f(z) is fixed at $z_0$. We continue it analytically along all curves starting at $z_0$ for which the continuation is possible. The resulting set of elements is called an analytic function generated by element f(z)

Here element f(z) at $z_0$ means it has power series expansion on some neighborhood of $z_0$

My question is:-

What is this analytic function generated by element f(z) formally? Is the analytic function generated by element f(z) over the complex plane a set or a function? It would also be good if anyone gives an example, like what is the analytic function generated by ln(z) ?

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    $\begingroup$ As you see there is nothing multivalued, just the set $\{ f_\gamma\}$ of all analytic continuations of $f$ along every curves $\gamma : z_0 \to ..$. Those analytic continuations glue together to form a Riemann surface. $\endgroup$
    – reuns
    Jul 3, 2019 at 18:21
  • $\begingroup$ This is a bit of a nuanced question depending on what you are asking. To add to reuns' comment, note that the behavior of different continuations differs from function to function... For example, we might have our natural domain be the entire unit circle in which case we get a bunch of copies of the same function. In the case of the logarithm, we get a bunch of spiraly branches, etc etc. Riemann Surfaces + sheaves are the classical theory I would start by exploring. $\endgroup$
    – Brevan Ellefsen
    Jul 3, 2019 at 18:28
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    $\begingroup$ Also note $\log$ is analytic $\Bbb{C}^* \to \Bbb{C/2i\pi Z}$ (where $\Bbb{C/2i\pi Z}$ is a Riemann surface) this time it is multivalued in the sense that $\log(z) = a+2i\pi \Bbb{Z}$ (and $z^{1/n} $ is analytic $\Bbb{C}^* \to \Bbb{C}^* / \langle e^{2i \pi /n}\rangle$) $\endgroup$
    – reuns
    Jul 3, 2019 at 18:29

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You're looking for the concept of complete analytic function. There are several ways to set up the details, but one of them is:

  • Let $\mathscr B$ be the set of all analytic functions defined on open discs in $\mathbb C$.

  • Define a relation $\sim$ on $\mathscr B$ such that $f\sim g$ iff the domains of $f$ and $g$ overlap, and their function values coincide on the intersection.

  • An anlytic continuation (which the above Wikipedia article calls a "global analytic function", though it doesn't have to be very "global") is any nonempty subset $\mathcal F$ of $\mathscr B$ such that every two $f,g\in\mathcal F$ can be connected with a finite chain of members of $\mathcal F$: $$ f \sim h_1 \sim h_2 \sim \cdots \sim h_n \sim g$$

  • A complete analytic function is an $\mathcal F$ that it maximal in that it cannot be extended without losing the above property.

  • Equivalently, a global analytic function is an equivalence class under the transitive closure of $\sim$.

For example, the global analytic function corresponding to the logarithm is the set of all functions that arise by (a) pick any open disc that doesn't contain the origin, and then (b) pick any branch of the logarithm on this disc, that is, let $f$ be such that $f'(z)=1/z$ everywhere and $e^{f(z)}=z$ at the center.

An analytic continuation in this sense is almost an atlas of its Riemann surface, except that the charts need to map $z$ not only to $f(z)$, but to a package containing both $z$ and all of the derivatives of $f$ at $z$ (or, equivalently, the coefficients of the power series at $z$) to avoid collapsing point on the Riemann surface.

It is not enough simply to construct a theory where a function can have multiple values, because the mere values of a function can coincide between different branches at a single point without anything else matching. For example consider $f(z)=(z-1)\sqrt z$ with the usual "multi-valued" square root. It has only one possible value at $z=1$, but its two branches nevertheless behave differently there -- they have different derivatives.

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  • $\begingroup$ So the global analytic function generated by f(z) at $z_0$ is same as global analytic function and it is a set and not a function is it so? $\endgroup$
    – Bijayan Ray
    Jul 5, 2019 at 4:32
  • $\begingroup$ @BijayanRay: Yes, in this representation. But you can equivalently represent it as a function that maps each $z$ to a set of different power series centered on $z$. Really it is an abstract thing that is neither really quite a set nor a function, but can be represented in either way. $\endgroup$
    – hmakholm left over Monica
    Jul 5, 2019 at 16:05

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