branch points of arcsin 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 $\displaystyle\frac{1}{\sqrt{1-x^2}}$ ? 
 A: 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. 
EDIT:
There is a long discussion about the branches of $\sqrt{z^2-1}$, following in the comment section, the argument boils down to the $\sqrt{z^2}$. Clearly
$$
\sqrt{z^2} = \mp z,
$$
this is why this function is multi valued. However,
$$
\sqrt{\left(e^{i0}\right)^2}=\sqrt{\left(e^{i2\pi}\right)^2}=1
$$
Or in other words, by going around the origin the function does not change value. Therefore, it is multi-valued but has no branch points!
This is the reason why $\sqrt{z^2-1}$ does not have a branch point at infinity, which can be checked by substituting $z=\frac{1}{t}$ and investigating $t\rightarrow 0$  .
A: 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 $\{z: 0 < |z - 1| < \epsilon\}$). 
