branch cuts / complex logs

**Squiggles** · October 10th 2008, 02:42 AM

Hi. So I'm new to branch cuts and I think I get the general idea, but I'm stuck with a couple homework questions along the same lines as:

I'm given the branch cut, eg (x=0, y>0)
and told for example log(-1) = -i(pi) on this branch
Then asked to find the value of some other simple complex logs on the same branch

I guess I'm just confused as to how the given value of log(-1) (or log(1) in another question) is relevent... do I need to use it to find a value of k in
log(z)=Log(r) + i(Argz + 2kpi) ?
but then I can't see how I would use the value of k to answer the problem... does the 2kpi even come into it once you have a branch cut? Am I looking at this totally wrong?? v. confused...

sorry if this isn't too coherent, i'm tired.. but any help appreciated!

**Opalg** · October 10th 2008, 03:14 AM

Remember that the imaginary part of the logarithm is i times the argument. If for example you are told that log(-1) = -iπ, it means that you are supposed to take -π (as opposed to +π or any other odd multiple of π) as the argument for -1. To stay on the same branch, you must ensure that the argument varies continuously as you move from -1 to any other point.

To take up your example: given the branch cut x=0, y≥0, and told that log(-1) = -iπ on this branch, suppose that you want to find log(1) on this branch. To get from -1 to 1 without crossing the cut, you have to go round below the origin. As you do so, the argument will increase from -π to 0, which means that you must take 0 as the logarithm of 1.

On the other hand, if (with that same branch cut) you were told to take log(-1) = iπ, that means that you must take π as the argument of -1. As you go continuously from -1 to 1, passing below the origin, the argument again increases, this time from π to 2π. So you would have to take log(1) = 2πi.

**shawsend** · October 10th 2008, 03:58 AM

First, you way better at this than me ok Opalg. However, I'd like to present a different approach to this in hopes it may be useful to others.

The key to understanding much about branch-cuts is the logarithmic multi-sheet. Get that and many of your problems with branch cuts are over. We know the log function is defined as:

$\log(z)=ln|r|+i \arg(z)$

What's the imaginary part of that function in terms of a real function of two real variables? It's $v(r,\theta)=\arg(z);\;z=re^{i\theta}$ with $\arg(z)=\text{Arg}(z)+2k\pi$ . What's a plot of that function look like? How about mine below? It's a helical sheet extending both up and down and radially without bounds although I've just clipped a small (r=3) portion of three turns of it. The overlapping of the sheets cause of course the multi-values since for any (x,y) pair, each sheet has a value. I've shown three values of $\text{Im}(\log(2-2i))$ as the blue, green, and red dot. However if we wish to define a single-valued analytic logarithm, we . . . cut . . . the multi-sheet so that we don't have any overlapping. I've color-coded three such non-overlapping ``branches'' of the log function in the figure. Each of these branches or sheets can be identified by:

$\begin{aligned}\log_0(z)&=ln|r|+i\Theta;\quad -\pi\leq\Theta<\pi;&\quad\text{(red)}\\<br /> \log_{-1}(z)&=ln|r|+i\Theta;\quad -3\pi\leq\Theta<-\pi;&\quad\text{(blue)}\\<br /> \log_{-2}(z)&=ln|r|+i\Theta;\quad \pi\leq\Theta<3\pi;&\quad\text{(green)}<br /> \end{aligned}$

Can you see why once I pick a branch, I can then identify a single value of $\log(z)$ for any z?

**ThePerfectHacker** · October 10th 2008, 05:40 AM

Originally Posted by Squiggles

Hi. So I'm new to branch cuts and I think I get the general idea, but I'm stuck with a couple homework questions along the same lines as:

I'm given the branch cut, eg (x=0, y>0)
and told for example log(-1) = -i(pi) on this branch
Then asked to find the value of some other simple complex logs on the same branch

Remember that to define a complex logarithm we need to pick a particular way for defining $\arg z$ since they are many ways.

The 'principal' logarithm is defined by $\log z = \ln |z| + i \arg z$ where $\arg z$ is the angle of the complex number given by $-\pi < \theta \leq \pi$ . For example, if we want to find $\log (-1)$ then we need to find $\theta$ which in this case is $\pi$ . Therefore, $\log (-1) = \ln |-1| + i \pi = \pi i$ . It turns out that $\log z$ is analytic everywhere except on $(-\infty,0]$ - this is called its branch. The reason is that the $\arg z$ function jumps values on this line, note it jumps from $-\pi$ to $\pi$ and vice-versa, thus it is not even continous there.

The logarithm along the positive axis is defined by $\arg z$ having values $0\leq \theta < 2\pi$ . Therefore, $\log (i) = \ln |i| + i \arg (i) = \tfrac{\pi}{2}i$ . It is exactly the same as above except $\theta$ is different. Compare what would have happened if we did the same computation using the principal logarithm, we would get a different result. Note that on $[0,\infty)$ this logarithm jumps values from $0$ to $2\pi$ . It turns out that this logarithm is analytic everywhere except on its branch.

The other thing we need to result is that if instead of defining $0\leq \theta < 2\pi$ we defined $0 < \theta \leq 2\pi$ it would not matter since $e^{i\phi} = e^{i(\phi + 2\pi)}$ . The point we need to realize here is that when defined a value for $\arg z$ it does not matter if we choose $(a,a+2\pi]$ or $[a,a+2\pi)$ since they gives the same function for $\arg z$ .

When the problem says to define logarithm along the branch $[0,i\infty)$ it is asking to define a logarithm so that $\log$ is analytic everywhere except on the line $[0,i\infty)$ . Looking at the above two examples it should be clear. We ought to define $\arg$ in such a way so that it jumps values from $a$ to $a+2\pi$ and vice-versa. If we pick $a=-\tfrac{3\pi}{2}$ it works. Therefore we define $\arg z$ by $\tfrac{-\pi}{2}\leq \theta < \tfrac{\pi}{2}$ . Therefore $\log (-1) = \ln |-1| + i \arg (-1) = -\pi i$ .

**Opalg** · October 10th 2008, 05:52 AM

Originally Posted by shawsend

Nice picture of a Riemann surface! It might be worth pointing out explicitly that what a branch cut does is to define the line that forms the boundary between one sheet of the helix and the next (differently-coloured) one. If you want to stay on the same branch then you have to keep to one of the coloured sheets.

**shawsend** · October 10th 2008, 08:12 AM

Here's something else I think can be helpful to readers struggling to understand branch cuts: Review the Mathematica code below and understand why it's generating the picture I made. Now, I created the cuts along the negative real axis. Change the code so that the branch cut is now along the positive imaginary axis (re-define the three branche functions $\log_0,\; \log_1,\;\log_{-1}$ corresponding to this change accordingly). Change the points so now the set is somewhere else. Generate a plot of the real component of $\log(z)$ . What's different about that plot? Plot both the real and imaginary components of $\log(a+bi)$ side by side (using GraphicsGrid[{{realplot,imagplot}}]). Just some ideas if any reader is interested. It helps me understand it and perhaps will you.

Code:

f[z_, k_] := Log[Abs[z]] + 
       I*(Arg[z] + 2*k*Pi); 
z1 = 2 - 2*I; 
point1 = Graphics3D[{{PointSize[0.04], 
           Green, Point[{{Re[z1], Im[z1], 
                 Im[f[z1, 0]]}}]}, {PointSize[0.04], 
           Blue, Point[{{Re[z1], Im[z1], 
                 Im[f[z1, 1]]}}]}, {PointSize[0.04], 
           Red, Point[{{Re[z1], Im[z1], 
                 Im[f[z1, -1]]}}]}}]; 

p1 = ParametricPlot3D[
     {Re[z], Im[z], Im[f[z, 0]]} /. 
       z -> r*Exp[I*t], {r, 0, 3}, 
     {t, -Pi, Pi}, PlotStyle -> Red]
p2 = ParametricPlot3D[
     {Re[z], Im[z], Im[f[z, -1]]} /. 
       z -> r*Exp[I*t], {r, 0, 3}, 
     {t, -Pi, Pi}, PlotStyle -> Blue]
p3 = ParametricPlot3D[
     {Re[z], Im[z], Im[f[z, 1]]} /. 
       z -> r*Exp[I*t], {r, 0, 3}, 
     {t, -Pi, Pi}, PlotStyle -> Green]
Show[{p1, p2, p3, point1}, 
   PlotRange -> {{-3, 3}, {-3, 3}, 
       {-3*Pi, 3*Pi}}, BoxRatios -> {1, 1, 1}]

**Squiggles** · October 10th 2008, 11:10 PM

thanks, that makes a lot more sense to me now! (well, enough to get by with my homework anyway

)
cheers

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