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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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