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I need help with this exercise:

For a suitable branch cuts we have that \begin{align*} \frac{d}{dz}\arcsin(z)=\frac{1}{(1-z^{2})^{1/2}}, \end{align*} for all $z\neq \pm 1$. Investigate the possible branch cuts that makes this formula valid.

Should I look at the branch point? Which should be at $\pm 1$ since the square-root is then $0$. But how should I know the branch cuts?

Or for the formula to be valid do I somehow have look at branch cuts on $\frac{d}{dz}\arcsin(z)$ and compare it to the branch cuts of $\frac{1}{(1-z^{2})^{1/2}}$?

Can someone please help me with this, I am abit lost. Thanks!

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    The plane can be cut with any "nice" contours that emanate from the branch points to the point at infinity. Examples include straight line paths from $\pm 1$ to $\pm \infty$ along the real axis. One can also show that the "slit" from $-1$ to $1$ is also a possible branch cut here. – 2017-02-22
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    So you mean that the answer is that possible branch cuts that makes this formula valid is $[-\infty, -1]$, $[-1,1]$ and $[1,\infty]$? How do I show this? – 2017-02-22
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    No, not quite. One possibility is $[-1,\infty)$ and $[1,\infty)$, which leads to $[-1,1]$ effectively. Another is $(-\infty,-1]$ and $[1,\infty)$. There are an infinite number of other possible cuts. – 2017-02-22
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    Thank you for your answer! But why is $(-\infty,1]$ not a branch cut then? I still don't quite get it. – 2017-02-22
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    That cut is actually two cuts, $1$ to $-\infty $ and $-1$ to $-\infty $. But, on the portion from $-1$ to $infty$, we can easily show that the function is single valued. – 2017-02-22
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    But isn't $1$ to $-\infty$ the same as $(-\infty,1]$? I'm still not sure on how I'm going to answer this question. It doesn't say much about branch cuts in my book and I have tried to find pages online but I find it hard. – 2017-02-23
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    Yes, it is the same. – 2017-02-23
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    But which cut do you mean is actually two cuts? Do you mean that $(-\infty,1]$ is two cuts? Another question, should it be $[$-paranteces or should it be $($-paranteces on both sides? – 2017-02-23
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    For the branch point at $1$, we could choose the cut to be from $1$ to $-\infty$. For the branch point at $-11$, we could choose the cut to be from $-11$ to $-\infty$. Note that these overlap from $-1$ to $-\infty$. In fact, on the overlapping portion, $f$ is single valued and hence continuous. And $f$ is analytic on $\mathbb{C}\setminus [-1,1]$. – 2017-02-23

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