How to calculate the Laurent series expansion of 1/(1-z)² in the region 1<|z|?

Eldonbetan
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1. I am trying to calculate the laurent series expansion of the function 1/(1-z)² in the region 1<|z|


2. None


3. I can get an answer informally by doing the polynomial division like in high school, but I don't know if this is the right answer and in case it is I cannot prove it. Any help would be largely appreciated.
 
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Recall that if a function is analytic on a domain, then its Laurent series and Taylor series are identical. Do you know how to do the Taylor series expansion?
 
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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