Detirmine where a laurent series is analytic

chanamn · March 26th, 2014, 02:08 AM

I have the following series and have to discuss where it is analytic
g(z)= sum_{n=1}^infinity ( i( n^2+3)/(3*(2n^2-1)))^n * (z^n + z^-n)

This is a laurent series. I found that without the negative powers of z, the series has a radius of converge of 1/6.
g(z) is abviously not analytic at z=0. But I thought the idea was to find some anulus where the whole thing is analytic.
How can I do this with z^-n part?
or is the answer as easy as we need |z|<1/6 and z not equal to zero, and in this case, how can I show this?

Thanks
chanman

HallsofIvy · March 26th, 2014, 07:03 AM

Since the $z^{-n}$ is the only "problem" and then only at z= 0, this is clearly analytic for all non-zero z. As an "annulus" that would be written as $0< |z|< \infty$ .

chanamn · March 26th, 2014, 11:51 AM

That is wrong to say because if the laurent series does not converge, then the function g(z) is not analytic

March 26th, 2014, 02:08 AM	#1
chanamn Newbie Joined: Mar 2014 Posts: 5 Thanks: 2	Detirmine where a laurent series is analytic I have the following series and have to discuss where it is analytic g(z)= sum_{n=1}^infinity ( i( n^2+3)/(3(2n^2-1)))^n (z^n + z^-n) This is a laurent series. I found that without the negative powers of z, the series has a radius of converge of 1/6. g(z) is abviously not analytic at z=0. But I thought the idea was to find some anulus where the whole thing is analytic. How can I do this with z^-n part? or is the answer as easy as we need \|z\|<1/6 and z not equal to zero, and in this case, how can I show this? Thanks chanman
$chanamn is offline$

March 26th, 2014, 07:03 AM	#2
HallsofIvy Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6	Re: Detirmine where a laurent series is analytic Since the $z^{-n}$ is the only "problem" and then only at z= 0, this is clearly analytic for all non-zero z. As an "annulus" that would be written as $0< \|z\|< \infty$ .
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March 26th, 2014, 11:51 AM	#3
chanamn Newbie Joined: Mar 2014 Posts: 5 Thanks: 2	Re: Detirmine where a laurent series is analytic That is wrong to say because if the laurent series does not converge, then the function g(z) is not analytic
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