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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: 2,531 Thanks: 1390 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: 2,531 Thanks: 1390 $z = (z-1) + 1$ $\dfrac{i}{z+1} =\dfrac{i}{(z-1)+2} = \dfrac i 2 \dfrac{1}{1+\frac{z-1}{2}}$ $\dfrac{1}{1+\frac z a} = \sum \limits_{k=0}^\infty~(-1)^k \left(\dfrac z a\right)^k$ so $\dfrac i 2 \dfrac{1}{1+\frac{z-1}{2}} = \dfrac i 2 \sum \limits_{k=0}^\infty~(-1)^k \left( \dfrac{z-1}{2} \right)^k$ and thus $z + \dfrac{i}{z+1} = 1 + (z-1)+\dfrac i 2 \sum \limits_{k=0}^\infty~(-1)^k \left( \dfrac{z-1}{2} \right)^k$ $z + \dfrac{i}{z+1} = \left(1+\dfrac i 2\right) + \left(1-\dfrac i 4\right)(z-1) + \dfrac i 2 \sum \limits_{k=2}^\infty~(-1)^k \left( \dfrac{z-1}{2} \right)^k$ Thanks from Country Boy and ghostwin
 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
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Quote:
 Originally Posted by ghostwin Thank you! And it possible to compute when z_{0} = -1?
just apply the same method

 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
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Quote:
 Originally Posted by romsek just apply the same method
There's a better method actually.

$z + \dfrac{i}{z+1} = \dfrac{i}{z+1} - 1 + (z+1)$

Thus the series about $z_0=-1$ consists of 3 terms

the -1st, 0th, and 1st.

the coefficients are seen to be

$c_{-1} = i$

$c_0 = -1$

$c_1 = 1$

 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!!

 Tags i or z, laurent, series

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