"Principal Branch Square Root of z in Domain C-{0}

ryou00730
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Homework Statement


Does the principal branch square root of z have a Laurent series expansion in the domain C-{0}?

The Attempt at a Solution



Well I'm not really sure what a principal branch is? I believe that there is a Laurent series expansion for z^(1/2) in C-{0} because originally our only problem is that when we take derivative of z^1/2 we get 1/z^[(2n+1)/2] and this is not defined at 0, but is everywhere else... so I think the answer is yes to this, but again I'm unsure of the details of principal branch?
 
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No. The square root is a multifunction and these functions do not have Laurent series about their branch-points because they're not fully analytic in a punctured disc surrounding the branch-point, specifically not so along their branch-cuts
 
But does taking the principal branch of square root z not deal with that? Does the principal branch mean we only take the principal roots of z?
 
The principal branch is analytic except along it's branch-cut which extends out from the origin so that we do not have an analytic domain in a punctured disk surrounding the origin and yes, the principal branch is the principal root with arg between -pi and pi
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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