Possible branch cuts for arcsin derivative

Tags:
1. Feb 14, 2017

ZuperPosition

1. The problem statement, all variables and given/known data
Our text book, Fundamentals of Complex Analysis, (....) by Saff Snider says on page 135 that by choosing some suitable branch for the square root and the logarithm then one can show that any such branch satisfies the equation below.

The homework/task is to find all such branch cuts such that this formula is valid.

2. Relevant equations

$$\frac{d}{dz}\sin^{-1}(z) = \frac{1}{(1-z^2)^{1/2}}$$ for all $z \neq 1,-1$, where $$\sin^{-1}(z) = -i \log(iz + (1-z^2)^{1/2})$$

3. The attempt at a solution

My initial hunch is that there is no such branch cut since any choice of branch for the square root and the logarithm have to imply that $\sin^{-1}(z)$ is not continuous for some value of the argument. Therefore the derivative of $\sin^{-1}(z)$ can not be well defined on all C except at z =1,-1?

Last edited: Feb 14, 2017
2. Feb 15, 2017

stevendaryl

Staff Emeritus
The existence of a branch cut just means that the function is multi-valued, not that it is not continuous. In the same way that $\sqrt{1} = \pm 1$, $sin^{-1}(1) = \pi \pm \frac{\pi}{2}$.

Multivalued functions are not a problem for real numbers--we can just say that we always choose the positive square root, and then it becomes single-valued. But in the complex plane, any choice you make runs into an inconsistency: If you start at $z=1$ and choose $\sqrt{z} = +1$, and then move around the circle in the complex plane with $|z| = 1$ and always pick the sign of the square root so that it is continuous, then when you get back to where you started, you will find $\sqrt{1} = -1$. So you cannot make $\sqrt{z}$ both single-valued and continuous everywhere. If you put in a branch cut, and say that the line along the real axis with $z > 0$ is not in the domain, then you can make the square-root single-valued and continuous everywhere else.

3. Feb 15, 2017

ZuperPosition

Thank you! Yes, I agree with what you are saying. So is it just sloppy notation by the book? That is, what they really want to say is that for some choice of branch cut the derivative is $$\frac{d}{dz}\sin^{-1}(z) = -i \log(iz + (1-z^2)^{1/2})$$ for all $z \in \mathbb{C}^*\backslash \{1,-1\}$ where $\mathbb{C}^*$ is the complex plane except the branch cut made? By this notation any branch cut should be ok, right?

4. Feb 16, 2017

stevendaryl

Staff Emeritus
Did you make a typo? It's not that the derivative is equal to that.

You can solve for the inverse sine as follows:

$y = sin^{-1}(z) \Rightarrow z = sin(y)$
$\Rightarrow z = \frac{e^{iy} - e^{-iy}}{2i}$
$\Rightarrow 2 i e^{iy} z = e^{2iy} - 1$
$\Rightarrow (e^{iy} - iz)^2 = 1-z^2$
$\Rightarrow e^{iy} = iz + \sqrt{1-z^2}$
$\Rightarrow y = \frac{1}{i} log(iz + \sqrt{1-z^2})$

So in relating $y$ to $z$, there are two choices to be made:
1. Which branch of the square-root?
2. Which branch of the log?
I'm not sure I understand the original problem: Are they asking for which choice makes the differential equation for $sin^{-1}$ valid? Or which branch makes the expression for $sin^{-1}$ in terms of $log$ valid?

5. Feb 22, 2017

ZuperPosition

Sorry, yes I wrote incorrectly, what I meant was $$\frac{d}{dz}\sin^{-1}(z) = \frac{1}{(1-z^2)^{1/2}}$$ for all $z$in $\mathbb{C}^*\backslash\{1,-1\}$.

Another way to restate the question is: Are there any branch cut such that $sin^{-1}(z)$ is holomorphic (simultaneously as being single valued) for all $z$ in $\mathbb{C}\backslash\{1,-1\}$?

6. Feb 22, 2017

stevendaryl

Staff Emeritus
We can show that $sin^{-1}(z)$ can't be single-valued and continuous everywhere.

Let $z = R e^{i \theta}$ and pick $R \gg 1$. Then we can write:

$dz = R i e^{i \theta} d \theta$

$sin^{-1}(z) = sin^{-1}(R) + \int_R^z \frac{1}{(1-z^2)^{1/2}} dz$

Let's pick an integration path parametrized by $\theta$, holding $R$ constant. Then we have:

$sin^{-1}(R e^{i\theta}) = sin^{-1}(R) + \int_0^\theta \frac{1}{(1-R^2e^{i 2 \theta})^{1/2}} R i e^{i \theta} d \theta$

Now, expand in powers of $\frac{1}{R}$:

$\frac{1}{(1-R^2e^{i 2 \theta})^{1/2}} = \pm i \frac{1}{R} e^{-i \theta} +$ higher order terms

So:

$sin^{-1}(R e^{i\theta}) = sin^{-1}(R) + \mp \int_0^\theta d \theta$ + higher order terms
$= sin^{-1}(R) + \mp \theta$ + higher order terms

So if we choose $\theta = 2\pi$, meaning we should get back to where we started, we find instead:

$sin^{-1}(R e^{i2\pi}) = sin^{-1}(R) \mp 2 \pi$

So $sin^{-1}$ changes by $2\pi$ when you go in a complete circle back to where you started. So it's not single-valued. So you have to have a cut.

Now, the question is: where to draw the cut? I'm not exactly sure, but the fact that we get into trouble when $z$ is large means that the cut has to go all the way out to infinity.

7. Feb 22, 2017

ZuperPosition

Oh, thank you so much! This definitely helps!