I am studying complex analysis and I have problem understanding the concept of branch cut.

The lecturer draw this as some curve that starts from a point and goes on and on in some direction (for example, something like $y=x$ for $x\geq0$ , but it doesn't have to be straight).

The definition given in the lecture is

A branch cut is a curve that is being presented in order to define a branch

and then he added a note

Points on the branch cut are singular.

Can someone please explain how does such a curve define a branch of a function ?

As I understand it, if $f(z)=u(r,\theta)+iv(r,\theta)$ then we want $\alpha<\theta\leq\alpha+2\pi$ for some real $\alpha$. How does a curve define this $\alpha$ ?

Why are the points on a branch cut singular? Are we also assuming something about the function that we are trying to define a branch of it?


A branch cut is something more general than a choice of a range for angles, which is just one way to fix a branch for the logarithm function. A branch cut is a minimal set of values so that the function considered can be consistently defined by analytic continuation on the complement of the branch cut. It does not alone define a branch, one must also fix the values of the function on some open set which the branch cut does not meet. Then analytic continuation uniquely defines a (single-valued) function on the complement of the branch cut, which function is a branch of the multi-valued function that analytic continuation without branch cut would define. By definition the points on the branch cut are singular in the sense that in their neighbourhood the limit of the values of the chosen branch does not exist; defining a value for the branch there in any way would make the branch discontinuous, so it is better left undefined.

Here is how it works for the logarithm. To start, the complex exponential function can be defined by $\exp z=\sum_{n=0}^\infty\frac{z^n}{n!}$, a series that converges for all $\def\C{\mathbf C}\def\i{\mathbf i}z\in\C$, and therefore determines an everywhere-defined single-valued function $\C\to\C$. The function is neither injective (since $\exp(z+2\pi\i)=\exp z$ for all $z$) nor surjective (sicne $\exp z\neq0$ for all $z$) so one cannot define the logarithm as its inverse function; however as a formal power series there is an inverse around $\exp0=1$ given by $$\ln(1+z)=-\sum_{n=1}^\infty\frac{(-z)^n}n.$$ This series is divergent for $|z|>1$ and for $z=-1$, but it defines a (single-valued) function $\ln:D\to\C$ for $D=\{\,z\in\C:|z-1|<1\,\}$, and it satifies $\exp(\ln z)=z$ for all $z\in D$. This function has no other singularities on the border of $D$ than at $z=0$, so its definition can be extended across the border at other points, by writing a power series around other points $z_0\neq1$ that coincide with the previous definition in the neighbouhood of $z_0$ but which continue to converge some way beyond $D$. Doing so, and still calling the extended function $\ln$, the equation $\exp(\ln z)=z$ continues to hold (the nature of the continuation process makes it impossible for such a relation to suddenly start to fail). So the proces defines an extended partial right-inverse of $\exp$. However, the continuation cannot be consistent when going around $0$ a full tour: the imaginary part of $\ln z$ must be a choice for the value of $\arg z$ (this follows from $\exp(\ln z)=z$), and this choice would increase by $2\pi$ when making such a tour. So in order to get a single-valued continuous function, one has to exclude some points of $\C$, other than the singularity at $0$ which is excluded by necessity, from the domain of $\ln$. This is where the choice of a branch-cut come in; in this case it must be a curve from $0$ to complex infinity, so that it cannot be avoided by any continuous path around the origin.

The usual choice for the branch cut is the negative real axis, giving the principal branch of the logarithm, but other branch cuts define other branches. Moreover, even with a given branch cut one could take another starting point than the value $\ln 1=0$; one could for instance take $\ln 1=6\pi\i$ instead and extend from there to the branch cut, or start in another point (which would even be necessary should $1$ lie on the branch cut), say with $\ln2\i=\ln2-\frac32\pi\i$. Since the relation $\exp(\ln z)=z$ continues to hold during continuation, one must require this relation on any branch that one defines, so one is not free to choose just any starting value for the logarithm in choosing a branch. However not all functions with branches are defined as the "inverse" of another function, so having such a guiding relation is somewhat particular for the case of the logarithm; the general rule for what can be the branch of an initially locally defined function, is that the branch can be obtained by repeatedly extending analytically and erasing old values of the function to allow them to be replaced with different values obtained by the continuation. For instance a branch of the logarithm with $\ln 1=6\pi\i$ can be obtained from our starting data above by continuing for three full counterclockwise tours around the origin.

  • $\begingroup$ Thanks for the answer! I am not that advanced in my study of complex analysis so I don't really understand your answer (for example I don't know what "analytic continuation" is). I am trying to understand how to use the curve to define a branch of a function and I have not manage to understand it from this answer. $\endgroup$ – Belgi Nov 27 '12 at 14:29
  • 1
    $\begingroup$ Analytic continuation is a general mechanism to extend the values of certain kinds of functions locally in the complex plane, as long as no singularities are encountered. Without this it makes no sense to talk about branches. Problems can arise globally when values to not match when continuing in different ways around singularities, for instance when going around $0$ for the logarithm. You then need to choose a "branch cut" to (artificially) stop this collision from happening. The values found up to but not including the branch cut constitute a branch. en.wikipedia.org/wiki/Branch_cut $\endgroup$ – Marc van Leeuwen Nov 27 '12 at 14:37
  • $\begingroup$ Wouldn't requiring continuity be enough to define the branch given one value and a brunch cut? $\endgroup$ – timur Dec 3 '12 at 7:49
  • $\begingroup$ @timur: If you are given a multi-valued function, then a branch cut and the choice of one of the multiple values at one point will (usually) determine the branch by continuity considerations alone. However in many cases the function is not born as a multi-valued function, but as a single-valued function defined in the neighbourhood of some point. If this function is analytic, then analytic continuation is the mechanism used to "push" the definition to a larger set (continuity is insufficient here); but doing so around a singularity may make the resulting "function" multi-valued. $\endgroup$ – Marc van Leeuwen Dec 3 '12 at 8:27
  • $\begingroup$ Thanks for the response. I was thinking in order to introduce a brunch cut in an appropriate way, one needs to already know something about the analytic continuation of the function, that is the whole picture. I am not sure how much information would be required though. $\endgroup$ – timur Dec 4 '12 at 1:25

I will demonstrate with the example of the complex logarithm.

Recall that a complex number $z=x+iy$ can be put into "polar form" $z=Re^{i \theta}$, where $R$ is the distance from $z$ to the origin and $\theta$ is the angle to $z$ around the complex plane measured positively from the $x$-axis. The complex logarithm has this formula for $z=Re^{i \theta}$:

$$\log z = ln |R| + i \theta.$$

You (hopefully) know that if $z=Re^{i \theta}$ then we can also represent $z$ as $z=Re^{i (\theta + 2\pi)}$ and generally as $z = Re^{i (\theta + 2 \pi k)}$. Consider what happens when you plug these different representations into the formula for $\log z$. If you use $Re^{i \theta}$ you get $\log z = ln |R| + i \theta$ but if you use $Re^{i (\theta + 8\pi)}$ you get $\log z = ln |R| + i (\theta + 8 \pi)$.

If you plot these points you run into an apparent problem -- they don't hit the same place! The consequence of this is that the complex logarithm is not a function (because functions take a single input to a single output -- this function takes a single input to multiple outputs).

Here is the "Riemann surface" for the complex logarithm:

$\qquad\qquad\qquad$ "Riemann surface" for the complex logarithm

-- don't worry too much about how it was generated but realize what it's telling you. If you pick the point $z = Re^{i \theta}$ on the complex plane and map it through the complex logarithm, we've seen you get these infinite number of different values; the picture just represents this.

What a branch cut does is restrict the outputs of the logarithm to one particular loop around this corkscrew surface (which one it is depends on the range of values you choose for your cut). Moreover, when we define branch cuts, there's a lot of points that we make "undefined" -- the reason for this is that while you can pick a particular loop to "go around" in the Riemann surface, you can never have continuity around the place you define the cut because the limits from different sides will be at different heights and so cannot be equal (and therefore it will not be continuous, which is generally a bad thing in calculus).

Let me give you something specific. For example, if you choose the branch cut to be the negative real axis, then you are allowing $-\pi < \theta < \pi$ -- that choice will roughly correspond to the "level" in the Riemann surface that has the red on it. Note that $z = Re^{-i \pi} = Re^{i \pi}$.

With this, you can see what I said earlier about continuity -- if you approach $Re^{i \pi}$ from values of $\theta$ greater than $\pi$, you would be on the yellow part of the surface but if you approach $Re^{i \pi}$ from values of $\theta$ less than $-\pi$, you are closer to the purple part of the image. We have approached $z = Re^{-i \pi} = Re^{i \pi}$ two different ways and got two different answers, in contradiction to continuity which we generally like to have in calculus.

edit: I'd like to add this domain coloring image of $\log z$ that I found in this Mathematica stack exchange post. The colors represent the argument of the output of the logarithm -- where it switches colors at angle $\pi$ is precisely the same jump discontinuity I'm talking about above:

$\qquad\qquad\qquad$ Domain colouring of $\log(z)$

  • $\begingroup$ Great motivating example, this clarified a lot for me. $\endgroup$ – AnotherPerson Aug 7 '16 at 2:06

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