# Analytic functions branch

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.

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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.

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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