All Laurent series expansion around 1.

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Homework Statement



Question is= Find all Laurent series expansion of f(z)=z^4/(3+z^2) around 1. I will be very very thankful if someone can help me to do this question.

Homework Equations





The Attempt at a Solution



can I assume (z-1=u) here and change the function in terms of $u$. then i will have

f(u+1)=(u+1)^4/(u+2)^2

It should have singularity at $u=-2$. now I m big confuse about in what regions I should compute Laurent series.

I have trouble of thinking that how many laurent series I will have and what will be the conditions on $z$.

Thanks !
 
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first, subbing in for z = u+1
g(u) = f(u+1)=(u+1)^4/(3+(u+1)^2) =(u+1)^4/(u^2+ 2u +4)
 
there will be a laurent series for each disk about z-1 = u = 0 defined by the poles of the function
 
You've done the substitution to center the Laurent expansion around u=0, now draw a diagram, plotting the poles of the new function and draw the annuli these poles split the plane into. For each annuli, decide what the rule is for that region (eg 8< |u| < 23 or something).

The next part is a bit harder: Try to manipulate the expression you have ( \frac{ (u+1)^4}{(u+1)^2+3} ) into the product of two terms, 1 a polynomial, and another that you can interpret as the sum of a geometric series. Then you can expand the geometric series into a sum, and multiply the polynomial into it to get the Laurent series.
 
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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