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November 2nd, 2017, 02:29 AM  #1 
Newbie Joined: Nov 2017 From: Portugal Posts: 6 Thanks: 0  Laurent Series of z + (i/z 1) when z = 1
Has anyone know how to solve the laurent series of z + (i / (z + 1)) when z = 1?

November 2nd, 2017, 03:02 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,103 Thanks: 1093 
when $z=1$ $z + \dfrac{i}{z+1} = 1+\dfrac{i}{2}$ i.e. is a complex number. Do you mean you want to find the laurent series expanded about $z_0=1$ ? 
November 2nd, 2017, 12:21 PM  #3 
Newbie Joined: Nov 2017 From: Portugal Posts: 6 Thanks: 0  Reply
Yeh, im sorry. I mean what you just said.

November 2nd, 2017, 12:52 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,103 Thanks: 1093 
$z = (z1) + 1$ $\dfrac{i}{z+1} =\dfrac{i}{(z1)+2} = \dfrac i 2 \dfrac{1}{1+\frac{z1}{2}}$ $\dfrac{1}{1+\frac z a} = \sum \limits_{k=0}^\infty~(1)^k \left(\dfrac z a\right)^k$ so $ \dfrac i 2 \dfrac{1}{1+\frac{z1}{2}} = \dfrac i 2 \sum \limits_{k=0}^\infty~(1)^k \left( \dfrac{z1}{2} \right)^k$ and thus $z + \dfrac{i}{z+1} = 1 + (z1)+\dfrac i 2 \sum \limits_{k=0}^\infty~(1)^k \left( \dfrac{z1}{2} \right)^k$ $z + \dfrac{i}{z+1} = \left(1+\dfrac i 2\right) + \left(1\dfrac i 4\right)(z1) + \dfrac i 2 \sum \limits_{k=2}^\infty~(1)^k \left( \dfrac{z1}{2} \right)^k$ 
November 2nd, 2017, 01:24 PM  #5 
Newbie Joined: Nov 2017 From: Portugal Posts: 6 Thanks: 0  Reply
Thank you! And it possible to compute when z_{0} = 1?

November 2nd, 2017, 02:09 PM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,103 Thanks: 1093  
November 2nd, 2017, 02:41 PM  #7 
Newbie Joined: Nov 2017 From: Portugal Posts: 6 Thanks: 0  Reply
How? $z = 1 \iff z + 1 = 0$ $w = z + 1$ so, $z + \frac{i}{z + 1} = w  1 + \frac{i}{w}$ And know? How to transform it in a geometric series? 
November 2nd, 2017, 02:50 PM  #8 
Senior Member Joined: Sep 2015 From: USA Posts: 2,103 Thanks: 1093  
November 2nd, 2017, 02:56 PM  #9 
Newbie Joined: Nov 2017 From: Portugal Posts: 6 Thanks: 0  Reply
Sorry for all my questions! But thank you so much!!


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i or z, laurent, series 
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