- #1

heman

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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.

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- Thread starter heman
- Start date

- #1

heman

- 361

- 0

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.

- #2

Hurkyl

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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)?

- #3

heman

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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..?

- #4

Hurkyl

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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...

- #5

heman

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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...

- #6

Hurkyl

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- #7

heman

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Hurkyl said:

"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!!

- #8

Hurkyl

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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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