# Branch cut

## Homework Statement

Given that $g(z) = ln(1-z^2)$, defined on $\mathbb{C}\backslash \left(-\infty, 1\right]$, i.e. the branch cut is from $-\infty$ to $1$ along the real axis. Find $g(-i)$ given $g(i) = ln(2)$.

## The Attempt at a Solution

I tried drawing it out but I'm having trouble making sense of this. I understand how $(-\infty, 0]$ works as a branch cut for $ln(z)$. If $ln(z) = a + bi$ we restrict $-\pi<b < \pi$ so there is a single value for $ln(z)$. This means every time $z$ passes through $(-\infty, 0]$ there is a discontinuous jump in the value of $ln(z)$.

However I've not been successful for explaining how $-\infty$ to $1$ works for $g(z) = ln(1-z^2)$.

From my drawing I think this means that the branch cut for $ln(g)$ is not $(-\infty, 0]$. But maybe that's not the right way to think about this.

It seems like if $\left| z\right|^2 > 1$, then making the argument of z go around a circle causes the argument of $1 - z^2$ to go around a circle twice.

vela
Staff Emeritus
Homework Helper
You can write the function as
$$g(z) = \log (1+z) + \log (1-z).$$ It has two branch points, one at z=1 and the other at z=-1.

I tried drawing it out but I'm having trouble making sense of this.

Yeah really. If you're just starting Complex Analysis then the best advice I can give you is to go out of your way to learn how to do this or you will never cultivate a deep understanding of multi-valued functions. We can always write the expression for the multi-valued log function as:
$$\log(g)=\ln|g|+i(\text{Arg}(g)+2n\pi),\quad n=0,\pm 1,\pm2\cdots$$
where $-\pi<Arg(1+z^2)\leq \pi$. So that in your case we have:
$$\log(1+z^2)=\ln|1+z^2|+i(\text{Arg}(1+z^2)+2n\pi),\quad n=0,\pm 1,\pm2\cdots$$
So what does that mean? Well, it means that the real part of the function is single-valued. It's just $\ln|1+z^2|$. However, the imaginary part is multivalued but more importantly, it's (locally) analytically continuous except at the branch points which are simply the points where $g=0$ and in your case, that's -1 and 1 (and the point at infinity). And that means there are no discontinuities except at the branch-points. Now, in practice, we excise a single-valued piece of this multi-valued function by creating entirely arbitrary "branch-cuts" that extend from one singular point (branch-point) to another singular point. But those branch-cuts are really NOT an integral component of the function, rather just artificial constructions to suit a particular need.

But let's get back to just plotting a part of this function pertaining to your question and since the difficulty lies with the multivalued-ness of the function, we'll plot the imaginary component. And the easiest way to do this is to define say three parts:

$$f0[z]:=\log(1+z^2)=\ln|1+z^2|+i(\text{Arg}(1+z^2)$$
$$fp1[z]:=\log(1+z^2)=\ln|1+z^2|+i(\text{Arg}(1+z^2)+2\pi)$$
$$fm1[z]:=\log(1+z^2)=\ln|1+z^2|+i(\text{Arg}(1+z^2)-2\pi)$$

And that's not at all difficult to plot in Mathematica:

Code:
reim = Im;
f0[z_] := Log[Abs[1 - z^2]] + I*Arg[1 - z^2];
fp1[z_] := Log[Abs[1 - z^2]] +
I*(Arg[1 - z^2] + 2*Pi);
fm1[z_] := Log[Abs[1 - z^2]] +
I*(Arg[1 - z^2] - 2*Pi);
p1 = ParametricPlot3D[Evaluate[
{Re[z], Im[z], reim[f0[z]]} /.
z -> r*Exp[I*t]], {r, 0, 2}, {t, -Pi, Pi},
PlotStyle -> Blue];
p2 = ParametricPlot3D[Evaluate[
{Re[z], Im[z], reim[fm1[z]]} /.
z -> r*Exp[I*t]], {r, 0, 2}, {t, -Pi, Pi},
PlotStyle -> Red];
p3 = ParametricPlot3D[Evaluate[
{Re[z], Im[z], reim[fp1[z]]} /.
z -> r*Exp[I*t]], {r, 0, 2}, {t, -Pi, Pi},
PlotStyle -> Green];
mypoint1 = Graphics3D[{PointSize[0.05], Black,
Point[{0, 1, reim[f0[I]]}]}];
Show[{p1, p2, p3, mypoint1}, PlotRange -> All,
BoxRatios -> {1, 1, 1.5}]

I should point out the plots have a spurious discontinuity at $[1,\infty)$ and is due to the plotting software. The actual multi-function is of course analytically continuous there. There are ways to make the plot smooth there but are involved.

And you'll notice I placed a black dot at $Im(f0)$. But this does not in itself answer your question. I've not created or rather, I've not excised a single-valued part of this multi-valued sheet with a branch-cut at $(\infty,1]$ such that it is analytically-continuous (and single-valued) everywhere except at the branch cut and having the property of $\text{mysheet}(i)=\ln(2).$. Can you do that? I'll give you a hint: half of blue, half of red, and get rid of green. :)

#### Attachments

• logfunction sheet.jpg
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Last edited:
vanhees71
Gold Member
2021 Award
You can write the function as
$$g(z) = \log (1+z) + \log (1-z).$$ It has two branch points, one at z=1 and the other at z=-1.

I'd be VERY careful following that suggestion. The function in the original problem is well defined, considering all the given data. In the complex plane you cannot so easily use
$$\ln(a b)=\ln a + \ln b \qquad (\text{WRONG!})$$
as for the real logarithm.

The original function is defined as given in the original problem with the branch cut as given. Of course you can also define the function with other cuts (one starting from $z=+1$ and one from $z=-1$), but that's a different function then!

vela
Staff Emeritus
Homework Helper
Yeah, I should've been more careful. The idea is to write the function as
$$g(z) = \log [-(z+1)(z-1)] = \log(-re^{i\theta}\rho e^{i\phi})$$ where ##z-(-1) = r^{i\theta}## and ##z-1 = \rho e^{i\phi}##. You want to determine appropriate ranges for ##\theta## and ##\phi## that are consistent with the branch cut and the other information given in the problem about g(z).

Well here is one way I was thinking about it.
$ln(-(z+1)(z-1)) = \pi + ln(z + 1) + ln(z - 1)$

Then for the first $ln$ on the right we can have a branch cut $\left(-\infty, -1 \right]$ and for the 2nd we have can have a branch cut $\left(-\infty, 1 \right]$. So the cut$\left(-\infty, 1 \right]$ works for the entire function.

However if what vanhees71 is wrote is right then maybe this is not a correct way. Based on vela's recent post we have

$ln(-(z+1)(z-1)) = ln(e^{i\pi}re^{i\theta}\rho e^{i\phi}) = ln(r\rho) + i\pi + i\theta + i\phi + i2\pi n$

But maybe it's not as clear what to do this way.

Last edited:
vela
Staff Emeritus
Homework Helper
You dropped a factor of ##i##. It should be ##\ln [-(z+1)(z-1)] = \ln (r\rho) + i(\pi + \theta+\phi+2\pi n)##.

The reason you don't want to say ##\log ab = \log a + \log b## is because ##\log## is a multivalued function, so the two sides could differ by a multiple of ##2\pi i##. The way I was thinking about the problem yesterday, everything worked out with n=0, so I overlooked the complication.

You're on the right track. Given the branch cut, what values would you assign to ##\theta## and ##\phi## when ##z=i##? What does ##n## have to equal so that ##g(i)## is purely real? Given that choice, you can then calculate ##g(-i)##.

You're on the right track. Given the branch cut, what values would you assign to ##\theta## and ##\phi## when ##z=i##?

I would assign them values of $\pi/4$ and $3\pi/4$. Although I may have a question about the "given the branch cut" part.

What does ##n## have to equal so that ##g(i)## is purely real? Given that choice, you can then calculate ##g(-i)##.

Then I suppose I would have to have $n = -1$ since the other imaginary terms add up to $2\pi i$.

Last edited:
So I ended up with

$g(-i) = ln(2) + i(-3\pi/4 - \pi/4 - \pi) = ln(2) - 2\pi i$