# $f(z)$ is holomorphic everywhere in the complex plane. It is positive for $z \in \mathbb{R}$. Can I choose $\log(f(z))$'s branch cuts to avoid $z=0$?

Suppose that $$f(z)$$ is holomorphic everywhere, and it's real and positive on the real axis (specifically, $$f(z) > \epsilon > 0$$ for all $$z \in \mathbb{R}$$).

I want to define an appropriate branch of $$\log(f(z))$$ so that $$\log(f(z))$$ is defined as the principal log on the real $$z$$ axis and is holomorphic everywhere in the complex plane except for its branch points (the locations where $$f(z)=0$$) and branch cuts emanating from those branch points.

Can I consistently choose the branch cuts of $$\log(f(z))$$ to avoid $$z=0$$? Specifically, I mean by "avoid" that if a branch point is distance $$R$$ away from the origin, the corresponding branch cut does not intersect any disk of radius $$r centered on the origin.

For example, maybe one could choose the branch cuts to always point radially outward towards infinity?

To clarify, the above definition of "avoids" can also be restated as "the branch cut emanating from a branch point does not pass closer to the origin than the branch point itself."

Here are a few examples that succeed at having all the branch cuts avoid the origin.

Here are a couple examples that don't work.

The main goal is to understand whether I can always choose the set of branch cuts of $$\log(f(z))$$ to never pass closer to the origin than the closest zero of $$f(z)$$, while also maintaining that $$\log(f(z))$$ is the usual real definition of log on the real $$z$$ axis, since $$f(z)$$ is real and positive there.

• Presumably, you want to avoid $f(z)=0,$ not $z=0?$ Commented Apr 19 at 17:27
• @ThomasAndrews I actually want to avoid $z=0$. Since $f(z)$ is positive for all real $z$, I'm wondering whether I can do that. Commented Apr 19 at 17:28
• @ThomasAndrews My analysis is a bit weak. Can I always choose branch cuts to point in any direction I choose? I was thinking there might be consistency constraints that restrict the way I draw branch cuts. If there aren't, then I will choose the branch cuts to point radially outwards. Commented Apr 19 at 17:32
• @ThomasAndrews My ultimate goal (which will be a separate question) is whether by forbidding branch cuts to approach $z=0$, the radius of convergence of the Taylor series of $\log(f(z))$ about $z=0$ is controlled by the location of the branch points -- if there are no branch points within a radius $R$, then the Taylor series for $\log(f(z))$ converges to $\log(f(z))$ within the disk of radius $R$ at the origin, and the Taylor series about $z=0$ does not converge at all for any point at distance $R'$ that is further from the origin than one of the branch points. Commented Apr 19 at 17:37
• I think you need to make branch cuts also connecting the zeros of $f$ to each other, besides connecting them to $\infty$. But I see no reason why you cannot avoid $z=0$ doing that. You can move the cuts freely as long as you stay inside the domain of holomorphy. Commented Apr 19 at 17:37

Some important definitions of branch cut for your question:

From Visual Complex Analysis by Needham. First Edition:

We draw and arbitrary (but not self-intersecting) curve C from the branch point $$z_0$$ out of to infinity; this is called a branch cut.

From Wolfram MathWorld:

A branch cut is a curve (with ends possibly open, closed, or half-open) in the complex plane across which an analytic multivalued function is discontinuous. For convenience, branch cuts are often taken as lines or line segments. Branch cuts (even those consisting of curves) are also known as cut lines or branch lines.

From MIT OpenCourseWare:

[Branch cuts] are curves joining the branch points in such a way as to prevent multiple values from arising (by eliminating paths that can go around the branch points). Then a single value can be selected (at each point in the cut plane) for the function. This then defines a branch of the multiple valued function.

We point out that the process by which a branch of a multiple valued function is constructed is somewhat arbitrary; the selection of the branch cuts is only limited by the fact that they must be there to prevent the multiple values from arising. This leaves a large amount of freedom in their selection. Thus, the branches of a multiple valued function are highly non-unique and (generally) there is nothing to make one branch special over another (excepting special reasons that may arise in specific applications and uses of complex variables).

Summarizing:

Branch cuts:

1. Are mostly arbitrary (not self-intersecting) curves.
2. Taken as lines or lines segment for convenience
3. The selection of the branch cut is only limited by the fact that they must prevent multiple values.
4. There is a large amount of freedom in their selection
5. Are highly non-unique
6. Nothing makes one branch special over another

So, for your case: $$\displaystyle \ln(f(z))$$

At first, if $$z=0$$ is not a zero of $$f$$ and not a branch point of $$\ln(f(z))$$, there seems to be nothing preventing you from constructing a branch cut that 'avoids' $$z=0$$. As pointed out in the definition, you can use arbitrary types of curves and paths.

However... the big warning: the level of vagueness in their construction should encourage you to deeply inspect the function $$f(z)$$ and not take a step forward without understanding its structure and properties. This is extremely necessary, as pointed out by Wolfram:

In general, branch cuts are not unique, but are instead chosen by convention to give simple analytic properties (Kahan 1987). Some functions have a relatively simple branch cut structure, while branch cuts for other functions are extremely complicated.

• +1 Thanks, this is good to read. I think I'm OK with a very complicated structure, so long as $\log(f(z))$ matches the usual principal log on the real axis. Do you see any impediments to that? (As an aside, my ultimate aim, which I've split into a few questions, is that I want to use a convergent series at the holomorphic point $z=0$ to argue that there are no branch points or other singularities near $z=0$ for my function of choice, and to do that I think I want to make sure that my branch cuts don't get close to $z=0$. I am most interested in the properties of $\log(f(z))$ on the real line.) Commented Apr 20 at 0:55
• @user196574 A set of necessary conditions for convergence of the McLaurin series of $\log(f(z))$ are: 1) $f$ should be analytic in an open disk $A_r$. 2) $\forall z\in A_r \; |\Im f(z)| <\pi$ Condition (2) varies with the branch cut of $\log(\cdot)$ selected. So, it is important the branch cut but also the image of the imaginary part of the function. Commented Apr 20 at 2:14
• However, it seems that you are more interested in sufficient conditions. Those are very hard: to prove that $\log(f(z))$ is analytical in a region $A$ using McLaurin for all $z_{0} \in A$, you must find an open disk $D_{r}(z_{0}) \subset A$ such that $\ln(f(z)) = \sum a_nz^n$ for all $z \in D_{r}(z_0)$. However, that works for the specific branch cut selected for $\log(\cdot)$; each branch cut generates a different function, and if you select a different cut, this may intersect with the discs $D_{r}$ Commented Apr 20 at 2:50
• Thanks. I know my function $\log(f(z))$ is holomorphic at the origin (inherited by the positivity of $f(z)$ on the real axis, and from the holomorphic-ness of $f(z)$), and I have a series for $\log(f(z))$. I also know the only singularities of $\log(f(z))$ are from branch points and branch cuts. I want to become confident that everywhere the series converges, it converges to $\log(f(z))$ (since I like the properties of the function described by the series). My thinking is that if I make sure my branch cuts "point away" from the origin, I'm safe. That inspired my question. Commented Apr 20 at 22:28