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From the definition given by wikipedia and Cauchy's theorem i can find the branch points of $\arcsin$ through its derivative $\displaystyle\frac{1}{\sqrt{1-x^2}}$

Are -1 and 1 simple pole of this expression ? (i'm a bit confused because of the fractional power)

Also, there is also a branch point at infinity. How do i find this branch point ? what are the order of all the branch points of arcsin ?

From wikipedia, i know that simple pole of derivative means logarithmic branch point, so there is no order if -1 and 1 are simple pole of $\displaystyle\frac{1}{\sqrt{1-x^2}}$ ?

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Just use the Euler formula $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$. Having $w=\arcsin(z)$ and $\sin(w)=z$ with a bit of algebra gives : $\arcsin(z)=-i\log(iz+(1-z^2)^{\frac{1}{2}})$. Just look at this. Because of the square root you have branch points at $z=\pm 1$, zero is not a branch point here. Infinity is a branch point because: If you substitute $z=\frac{1}{t}$ and look what is going on if $t \to 0$ you would enevetably convince yourself that infinity is a brnch point. i.e you have term of the sort $\log(t)$. So you should have branch cut extending to infinity.

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No, branch points and poles are different. A pole is an isolated singularity, a branch point is not (e.g. there is no way to define $\arcsin(z)$ as an analytic function in a punctured disk $\{z: 0 < |z - 1| < \epsilon\}$).

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    You can look at the series expansions of $f'(x) = 1/\sqrt{1-x^2}$ around each of its singularities. Around $x = 1$, for example, the expansion of $f'(x)$ (and therefore also of $f(x)$) involves half-integer powers of $x-1$. Around $x=\infty$, $f'(x) = \pm (i/x + i/(2x^3) + \ldots)$, and the term in $1/x$ gives $f(x)$ a logarithmic term.2012-11-21