Laurent series of the function f(z)=1/z^2 for |z-a|>|a|

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Hi all,,

I have an Exam tommorow and this question is irritating me...Pls help

Laurent series of the function f(z)=1/z^2 for |z-a|>|a| .a is not equal to zero...
I am waiting for yours responses...I will be highly thankful to you.
 

Hurkyl

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Don't you have a formula for the coefficients, that involves integrating over the annulus |z - a| > |a|?

Or, maybe you could look at it as:

|1 / (z-a)| < 1/|a|

And try to rewrite 1/z^2 so it looks like the sum of a geometric series in 1/(z-a)?
 
346
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Omg,,,I did not use the formula for the coefficients,it did not seemed to strike me,,,yeah then its easy...and can be solved...

I was trying to do it by bringing in the form of geometric series..
Hurkyl,
Pls tell me how can i write 1/z^2 in that apt form as here the power of z is 2....
I seem to moving forward but then 2 series will be multiplied acc. to me,won't it be tedious..?
 

Hurkyl

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Hrm, yah, writing it as a geometric series won't work... but maybe the sum of two geometric series?

I guess it would be easier to observe that d/dz (-1/z) = 1/z^2, so if you had a Laurent series for -1/z...
 
346
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Oh yes,,,i got it ....That will work....Thanks Hurkyl ,I feel much better now!!

But in writing the Laurentz Series for F(z)=exp(z+1/z) around zero ,i think i have got no escape ...i think i am bound to multiply the series for e^z and e^(1/z),,,But how will i write the coefficient for each term,,or there can be any better approach...
 

Hurkyl

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I don't see the difficulty in multiplying the series... I guess you don't like the fact that each coefficient will be an infinite sum. :smile: (Though, it wouldn't surprise me if there's a clever way to do this that gives you a closed form)
 
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Hurkyl said:
I don't see the difficulty in multiplying the series... I guess you don't like the fact that each coefficient will be an infinite sum. :smile: (Though, it wouldn't surprise me if there's a clever way to do this that gives you a closed form)

"Each coefficient will be an infinite sum."....that i can see but where it will start that seems out of my reach.....limits are worrying me...i am able to see it mechanically only...
I think i sincerely need help on this!!
 

Hurkyl

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Have you taken an analysis or advanced calculus course? The relevant theorems should be in there.

The main thing here, though, is that the series involved are absolutely convergent, meaning that you can rearrange terms in any way you please.
 

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