Calculating the residue of a complex function

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The discussion revolves around calculating the residue of the function f(z) = (1 + z^2)^-3 at the pole z = i, which is believed to be of order 3. Participants clarify that to find the residue, one must use the limit formula involving the order of the pole, specifically lim(z→a) (z-a)^3 f(z). There is a focus on the importance of factorizing the denominator to simplify the calculation. The initial attempt yielded a result of 3/8, but it was noted that additional factors were likely missing in the calculation. The conversation emphasizes the need for careful application of the residue theorem and proper limit evaluation to accurately determine the residue.
AlBell
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Homework Statement



calculate the residue of the pole at z=i of the function

f(z)=(1+z^2)^-3

State the order of the pole

Homework Equations



I know the residue theorem and also the laurent series expansion but I'm having trouble applying these

The Attempt at a Solution



I think the pole is order 3 but I'm having trouble either expanding the function in order to find the a(-1) term or implementing the residue theorem and would appreciate some help! Thanks
 
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Do you know that the residue of a function f at the point a can be calculated by

\lim_{z\rightarrow a} (z-a)f(z)
 
I thought it was
dxzktg.png
?
 
AlBell said:
I thought it was
dxzktg.png
?

Yes, of course, sorry. My formula is for a simple pole.
So, can you use your formula to find the residue?
 
Yes but I am having trouble actually implementing the formula and wondered if anyone could do an example to help me understand it?
 
Well, first you'll need to determine the order of the pole.

Your guess is that the order of the pole is 3. So you'll need to look at

\lim_{z\rightarrow a} (z-a)^3f(z).

This limit should be nonzero for the order of the pole to be 3. If it is zero, then the pole is of a lower order. If the limit doesn't exist, then the pole is of higher order.

So, can you calculate that limit?

Hint: factor the denominator.
 
Ah I forgot about factorising, that makes it so much simpler! I get 3/8 I think!
 
VERY close. You probably got something like 12*(2i)^(-5), right? Just compute this again. The 3/8 is right, but you are missing two factors.
 

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