Laurent Series for f(z)=(1+2z)/(z^2+z^3)

opticaltempest
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Would anyone be willing to check and comment on my work for finding the Laurent series of

f(z)=\frac{1+2z}{z^2+z^3} ?

Page 1 - http://img23.imageshack.us/img23/7172/i0001.jpg"

Page 2 - http://img5.imageshack.us/img5/2140/i0002.jpg"

Page 3 - http://img15.imageshack.us/img15/2753/i0003.jpg"

I also displayed the pages below.

Page 1

http://img23.imageshack.us/img23/7172/i0001.jpg


Page 2

http://img5.imageshack.us/img5/2140/i0002.jpg


Page 3

http://img15.imageshack.us/img15/2753/i0003.jpg


Thanks in advance!
 
Last edited by a moderator:
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It looks good to me!
 
Thanks!
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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...
View full post »

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