Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am having trouble understanding branch cuts. It seems right when I understand one thing another issue arises. The questions asks:

Find a branch of each of the following multiple valued functions that is analytic in the given domain:

a.) $(z^2-1)^{1/2}$ in the unit disk, $|z|< 1$.

b.)$(4+z^2)^{1/2}$ in the complex plane slit along the imaginary axis from $-2i$ to $2i$.

c.) $(z^4 -1)^{1/2}$ in the exterior of the unit circle $|z|>1$.

d.) $(z^3-1)^{1/3}$ in the exterior of the unit circle, $|z|>1$.

For a, I had a question earlier similar to this and it says the prinicipal branch won't work because it is not defined in some of the x-axis and all of the y-axis (not entirely sure why that is though). The answer to this question is $ie^{[(1/2)Log(1-z^2)]}$, which doesn't make sense to me because it says the principal branch is not defined there.

For b, I kind of do understand because if it is on the imaginary axis from $-2i$ to $2i$ which will be undefined so I must use another branch.

For c and d both similar to a.

share|improve this question

1 Answer 1

up vote 9 down vote accepted

Remember that the principal branch of $\log$ (which your book denotes by Log) is defined everywhere except for nonpositive real numbers.

For a: When your book says "the principal branch won't work", it means that the "obvious" answer of $\exp(\frac{1}{2} Log(z^2-1))$ isn't correct. This is a branch of $(z^2-1)^{1/2}$, but isn't defined everywhere on $|z|<1$. That's because at some points in $|z|<1$, the argument $z^2-1$ to Log is a nonpositive real number, where Log is not defined. Where does this happen? We have $z^2-1 \le 0$ iff $z^2 \le 1$. That naturally splits into 2 pieces: If $0 \le z^2 \le 1$, then $z$ will be in the real interval (-1,1) along the real axis, but if $z^2 \le 0$, then $z$ can be anywhere on the imaginary axis. So Log($z^2-1$) will fail on all those points.

However, if we use the function Log($1-z^2$) instead, then that is defined everywhere in the domain $|z|<1$. That's because $1-z^2$ cannot be a nonpositive real number if $|z|<1$ (check this for yourself). Now we can write $\exp(\frac{1}{2}Log(1-z^2))$, which is defined everywhere in $|z|<1$. That gives us a branch of $(1-z^2)^{1/2}$, which is close to, but not quite, what we want. We want a branch of $(z^2-1)^{1/2}$. So, we multiply a value of $(-1)^{1/2}$, which could be either $i$ or $-i$. Thus we arrive at the answer you show above: $i \exp(\frac{1}{2}Log(1-z^2))$ (multiplying by $-i$ instead of $i$ would also be correct; the negative of a branch of a square root is also a branch).

To check that this answer gives you a branch of $(z^2-1)^{1/2}$, you should square it and see that the result is $z^2-1$.

How might one come up with the idea to try Log($1-z^2$)? Here's one way. First, notice that our problem will be solved if we define a branch of $\log(z^2-1)$ in $|z|<1$, because once we do that we can write $\exp(\frac{1}{2} \log(z^2-1))$. However, we cannot choose the principal branch Log for log, for reasons we have already seen.

Let's look at the reason in a little more detail. The region $|z|<1$ is a unit disc centered at 0. What happens if we apply $z^2-1$ to it? If we square all points of the unit disc centered at 0, we still get the same unit disc. Then if we subtract 1, we get a unit disc centered at -1, which I will call $D$. Because $D$ contains nonpositive real numbers, the principal branch Log doesn't work. However, if we were to multiply $D$ by -1, we would get a unit disc centered at 1. That disc doesn't contain any nonpositive real numbers and therefore Log is okay. This suggests using $-(z^2-1) = 1-z^2$ as the argument for Log.

One more point: Negating the argument is a relatively harmless thing to do for logarithms because of the "identity" $$\log(-z) = \log(z) + \log(-1).$$ Here I put "identity" in quotes because this "identity", extrapolated from the true identity for real logarithms $\log(xy) = \log(x) + \log(y)$, isn't quite right for complex logarithms. You can't actually choose a branch for which the identity holds everywhere. Nevertheless, it suggests that, for the original problem, negating the argument $z^2-1$ to $1-z^2$ is a relatively harmless thing to do and we'll be able to recover by a relatively simple operation of adding $\log(-1)$. (In the actual solution, we multiplied by $(-1)^{1/2}$, which is what happens after you apply the exponential: $\exp(\frac{1}{2} \log(-1)) = (-1)^{1/2}$.)

I hope this helps you proceed similarly for the other 3 parts.

share|improve this answer
    
I hope this gets a lot more votes up!! I am very grateful for your thorough explaination! Thank you so much, Ted!!! –  Q.matin Jan 29 '13 at 7:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.