Calculation residues at z=1 of order 4

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Homework Statement



Find the residue of the following function at z=1

Homework Equations


f(z)=z^2/(z-1)^4(z-2)(z-3)


The Attempt at a Solution


lim z→2 z^2/(z-1)^4(z-3)=-4

limz→3 z^2/(z-1)^4(z-2)=9/16

at z=1 of order 4?? using formula 1/m-1! d/dz{(z-a)^m f(z)}
 
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Kowsi Ram said:

Homework Statement



Find the residue of the following function at z=1

Homework Equations


f(z)=z^2/(z-1)^4(z-2)(z-3)


The Attempt at a Solution


lim z→2 z^2/(z-1)^4(z-3)=-4

limz→3 z^2/(z-1)^4(z-2)=9/16

at z=1 of order 4?? using formula 1/m-1! d/dz{(z-a)^m f(z)}

Looks good, except that you need to take m-1 derivatives (why?).
 
Thanks for your reply. Please can you solve it and show me, sir.
 
I think it would be better if you did your homework yourself.

Remember the definition of a residue -- there's an easier way to calculate it in this case, than differentiating the function three times. Expand it into a Laurent series around z=1 by multiplying a couple of geometric series together.
 
You are correct. I will try this method.
 
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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