2
$\begingroup$

I am studying complex analysis and I do not yet fully understand branch points and branch cuts. I am trying to figure out how it works by looking at the following:

$z \rightarrow \frac{1}{2i} \log(\frac{1+iz}{1-iz})$

(arctan(z)= $\log(\frac{1+iz}{1-iz})$ )

Now how do I find the branch point and branch cuts?

$\endgroup$
5
  • 1
    $\begingroup$ Just a passing remark: If $f$ is a (multi-valued) function, the branch points of $f$ (points $p$ in the domain such that $f$ is multi-valued in every neighborhood of $p$) are determined by $f$, but the branch cuts of $f$ (arcs that are removed from the domain of $f$ in order to get a continuous branch of $f$) are not determined by $f$. Strictly speaking, you can choose branch cuts, but not find them (in the usual mathematical sense of "find"). $\endgroup$ Commented Oct 12, 2013 at 16:17
  • $\begingroup$ @AndrewD.Hwang How does it work to choose suitable branch cuts in this case? $\endgroup$
    – student441
    Commented Oct 12, 2013 at 18:08
  • $\begingroup$ There is no (continuous) branch of $\log$ on any punctured neighborhood of $0$ or $\infty$, but there is a branch of $\log$ on every simply-connected subset of the set of non-zero complex numbers. (As you probably know, the conventional choice is to remove $(-\infty,0]$, and to take $\log z$ real on the positive real axis.) The branch points of $\arctan$ are the $z$ for which $(1+iz)/(1-iz)=0$ or $\infty$. Remove an arc joining these points to obtain a region on which $\arctan$ has an analytic branch. Jonathan Y.'s hint tells you how to ensure $(1+iz)/(1-iz)$ is non-positive real. $\endgroup$ Commented Oct 12, 2013 at 20:19
  • $\begingroup$ @AndrewD.Hwang How can I ''see'' that there is no branch of log on any punctured neighborhood of $0$ or $\infty$? And then subsequently that there is a branch of log on every simply-connected subset of the set of non-zero complex numbers? $\endgroup$
    – student441
    Commented Oct 13, 2013 at 9:05
  • $\begingroup$ @AndrewD.Hwang With regard to your first (helpful) remark, I am also bothered what it intuitively means to ''have a continuous branch''. Sorry for asking all these questions; this subject confuses me a lot... $\endgroup$
    – student441
    Commented Oct 13, 2013 at 9:25

2 Answers 2

2
$\begingroup$

Hint: for every $a,b\in\mathbb{C}$, find those $z\in\mathbb{C}$ such that $\frac{z-a}{z-b}\in\mathbb{R}$. Then find exactly when the given fraction is positive, and when it's negative.

Edit: for any $u,w\in\mathbb{C}$ we have $\arg{\frac{u}{w}} = \arg{u}-\arg{w}$. It follows that the quotient is real iff $u,w$ have the same argument (they 'point' in the same direction) or opposite arguments. How can we apply that to the first hint?


Edit: the next part is more explicit, and I'll try to give some intuition based on your familiarity with Möbius transformation (however, if you're not done thinking about the problem, you might wish to delay reading it). As mentioned by @AndrewD.Hwang in the comments above, an analytic branch of logarithm exists in any simply connected domain not including zero. Put differently, when you think of the complex plane as the Riemann sphere (infinity as the 'north' pole), the logarithm has branch points at the poles (zero and infinity), and removing any arc connecting both poles (that arc becomes the branch cut) will yield a simply connected surface on which an analytic branch of logarithm indeed exists.

Now, $\varphi(z) = \frac{1+iz}{1-iz}$ is a Möbius transformation. This means that it's conformal on the entire Riemann sphere (if you'd like, it 'distorts' the sphere in general, but locally it behaves very similarly to shifts and rotations). Note that it maps $i\mapsto 0$, $-i\mapsto \infty$, implying that the image of any arc connecting $\pm i$ under $\varphi$ is an arc connecting $0,\infty$, and indeed $\pm i$ are the branch points of $\arctan := \log\circ\varphi$, and any arc connecting them would be a branch cut (in other words, removing such an arc would yield a domain in which an analytic branch of $\arctan$ exists).

Now, as AndrewD.Hwang also mentioned, the standard 'choice' of branch cut for logarithm is the non-positive real line. It is here that my original hint could help us, as it allows us to find the arc mapped by $\varphi$ to that cut. As I said, for any $a,b\in\mathbb{C}$ we have $\frac{z-a}{z-b}\in\mathbb{R}$ iff $(z-a),(z-b)$ point in similar or opposite directions. This happens exactly when $z$ lies on the unique line passing through $a,b$, and the quotient is negative iff $z$ lies on the segment connecting $a,b$ (because that is when the arrows from $a,b$ to $z$ point in opposite directions). More rigorously, one notes: $$\frac{z-a}{z-b}=t \iff z-a=t(z-b) \iff (1-t)z = a - tb = (1-t)a - t(b-a)\\ \iff z = a + \frac{t}{t-1}(b-a),$$ and indeed $\frac{t}{t-1}\in(0,1)$ ($z$ lies between $a,b$) iff $t$ is non-positive.

Finally, note that $$\varphi(z) = \frac{i(z-i)}{i(-i-z)} = -\frac{z-i}{z+i}$$ is a non-positive real exactly when $z$ lies on the imaginary axis but not between $\pm i$, i.e. on $$\{it\mid t\in\mathbb{R}, |t|\geq 1\}.$$ (This, you see, is an arc in the Riemann sphere connecting $\pm i$, which has the 'nice' property of passing through infinity--the point on the sphere that we 'dislike' to work with. Or, if you'd like, it's the only arc connecting $\pm i$ which doesn't pass through the real line, allowing us to truly extend the familiar $\arctan$ on reals.)

$\endgroup$
13
  • $\begingroup$ Can you further clarify it? I can't make anything of it unfortunately... $\endgroup$
    – student441
    Commented Oct 12, 2013 at 13:13
  • $\begingroup$ @student441, do you need help solving the hint, or applying it to the problem at hand once it's solved? $\endgroup$ Commented Oct 12, 2013 at 13:18
  • $\begingroup$ Both actually :'). I am totally new to complex analysis and I am really stuck on this topic right now.. Your help would be very much appreciated. $\endgroup$
    – student441
    Commented Oct 12, 2013 at 13:20
  • $\begingroup$ @student441 that's okay. Let focus on the hint first; the rest might become apparent once we've done that. I'll edit. $\endgroup$ Commented Oct 12, 2013 at 13:22
  • $\begingroup$ I get from your edit that: z-a and z-b should point in the same or opposite direction. Is that correct? $\endgroup$
    – student441
    Commented Oct 12, 2013 at 13:32
1
$\begingroup$

(This answers questions that arose in comments, but is itself too lengthy for a comment.)

If $f:X \to Y$ is a mapping, a "branch of $f^{-1}$" is a mapping $g:Y \to X$ such that $(f \circ g)(y) = y$ for all $y$ in $Y$, a.k.a. a right inverse of $f$. (Note carefully that $(g \circ f)(x) = x$ does not generally hold for all $x$ in $X$, but only for $x$ in the image of $g$. Think of the positive branch of the real square root, or the inverse trig functions.)

In particular, if $U$ is a (usually open) set of complex numbers, a branch of $\log$ in $U$ is a mapping $L:U \to \mathbf{C}$ such that $\exp\bigl(L(w)\bigr) = w$ for all $w$ in $U$. If $L$ is continuous, we call $L$ a continuous branch of $\log$, and say there exists a continuous branch of $\log$ in $U$.

To address the questions of (i) how to see there is no continuous branch of $\log$ on any punctured neighborhood of $0$ or $\infty$ and (ii) there is such a branch near an arbitrary non-zero number, here's a depiction of the complex logarithm:

Riemann surface of log

Writing $z = x + iy = re^{i\theta}$ (with $r>0$, and $x$, $y$, and $\theta$ real) and $w = \log z = (\log r) + i\theta$, the surface has equation $Z = \operatorname{Re}(w) + \operatorname{Im}(w) = (\log r) + \theta$ (with $Z$ denoting the height coordinate in $\mathbf{R}^3$).

Geometrically, this surface is a "helicoid" swept out by the graph $\operatorname{Re}(w) = \log x$ by simultaneously revolving at unit angular speed about the $Z$ axis and translating at unit speed along the $Z$ axis. It consists of infinitely many "sheets" (three of which are shown), joined one to the next. A branch cut is the image $\gamma$ of a piecewise $C^1$ curve in $\mathbf{C}$ joining $0$ to $\infty$. If you remove points on the surface lying above $\gamma$, the surface separates into infinitely many graphs, each the geometric representation of a continuous branch of $\log$ on $U = \mathbf{C} \setminus \gamma$.

The green curve is a "lifting" of two full turns of the unit circle in the $z$ plane; the fact that the lift is not a closed curve manifests the fact that there is no continuous branch of $\log$ in a punctured neighborhood of $0$. The picture near $\infty$ is congruent, since "$\log(1/z) = -\log z$", with the understanding that each side is multi-valued.

$\endgroup$
1
  • $\begingroup$ $\arctan=\frac{i}{2}\left( \frac{i+z}{i-z}\right)$ and $arccot=\frac{i}{2}\left( \frac{z-i}{z+i}\right)$. Both of these have branch point at $-1,1$. From what I can see the branch cuts can be either $(-1,1)$ or $(-\infty,-1) \cup (1,\infty)$. Depending on wether we connect $-1,1$ via infinity. Can $\arctan$ and $arccot$ really have the same branch cuts? $\endgroup$ Commented May 18, 2020 at 19:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .