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I found one author who said that complex derivative of

$$\sin^{-1}z=-i\log(zi+(1-z^2)^{1/2})$$

is the same on all branches, but I found another who says they can differ. Can anyone share an example of where the derivative might differ on different branches?

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2 Answers 2

Given a branch $f$ of $\arcsin$, there will be other branches of the form $f(z) + 2k\pi$, and also branches of the form $(2k+1)\pi - f(z)$. The former will have the same derivative as the original, but the latter will have derivative equal to the negative of the original branch's derivative.

As suggested in @mrf's answer, we should compare this with $\log z$ or $\arctan z$. All the branches of $\log z$ (respectively, $\arctan z$) are obtained by adding multiples of $2 \pi i$ (respectively, $\pi$) so their respective derivatives will not have branch points.

Also consider $\sqrt{z}$. Negating a branch gives another branch, so the derivatives will be negatives of one another.

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The derivative is $$\frac{1}{(1-z^2)^{1/2}}$$ so each choice of branch for $(1-z^2)^{1/2}$ (which is needed to define $f(z) = \arcsin z$) gives a choice of branch for $f'$.

Compare this to $\log z$ or $\arctan z$ where the derivatives don't have branch points even though the functions themselves do.

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