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#1 2015-10-25 20:05:37

Maximoff
Member
Registered: 2015-10-23
Posts: 10

The Residue Theorem

Hello everyone. Could anybody help me with this? Its about residue theorem.

Find the residue at z=i of the function

Thanks in advance.

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#2 2015-10-26 18:23:32

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: The Residue Theorem

Hi;

is what I am getting.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#3 2015-10-26 22:09:52

Maximoff
Member
Registered: 2015-10-23
Posts: 10

Re: The Residue Theorem

Hi bobbym..

Could you show the working step on how you get it? I got a less bit of knowledge on this theorem.
Thanks.

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#4 2015-10-27 00:19:21

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: The Residue Theorem

I used a CAS on that.

What methods have you been taught for simple poles?


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#5 2015-10-27 05:07:39

zetafunc
Moderator
Registered: 2014-05-21
Posts: 2,432
Website

Re: The Residue Theorem

Firstly, note that

so there is indeed a pole at z = i, and it has multiplicity two. Recall that, if a is a pole of multiplicity n, then the residue of f around the pole z = a is given by the following formula:

Notice that for a simple pole (i.e. n = 1) we can drop the differential operator. However, in this case, the pole is not simple.

Now we just put n = 2, and a = i. We obtain the following:

where the last equality is just algebra, from plugging in z = i.

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#6 2015-10-27 05:13:23

zetafunc
Moderator
Registered: 2014-05-21
Posts: 2,432
Website

Re: The Residue Theorem

You also asked a question on the LearnMathsFree Facebook page, which I will respond to here:

To deal with
, we can first multiply the top and bottom by
, to get:

, which you can then turn into the form z = x + iy by multiplying by the conjugate of the denominator in the usual way.

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