What is the residue at ##z=0## for ##\frac{1}{z^3}+e^{2z}##?

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Homework Statement



Find the residue of ##\frac{1-cos2z}{z^3}## at ##z=0##

Homework Equations



##Res=\frac{1}{n!}\frac{d^n}{dz^n}[f(z)(z-z_0)^{n+1}]## Where the order of the pole is ##n+1##

The Attempt at a Solution



Differentiating ##(1-cos2z)z^3## twice, leaves me with zeros against every term giving me ##0## for a residue which is incorrect. Mathematica gives me 2 for a residue. Since ##z_0=0## I am only left multiplying the top function by z. I have no idea what I am missing here.

What am I doing incorrectly?

Thanks
Chris
 
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What is the order of your pole?
 
3rd order. I think I made a mistake that the 1/z^3 factor was not included in f(x). If I include it, it cancels out the z's outside of the cos2z function altogether. In that case, the part that gets differentiated and evaluated is just 1-cos2x.

Thanks,
Chris
 
kq6up said:
3rd order.
It is not. Don't forget the numerator.
 
I got it. I having trouble with a different problem now. A new thread is in order for that one.

Thanks,
Chris
 
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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