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 Complex Analysis Complex Analysis Math Forum

 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 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Robotboyx9 Calculus 2 October 25th, 2017 08:07 AM yusuf zeyd Complex Analysis 0 May 17th, 2015 02:37 AM king.oslo Complex Analysis 0 December 28th, 2014 06:50 AM WWRtelescoping Complex Analysis 7 May 5th, 2014 11:09 PM capea Complex Analysis 1 October 25th, 2011 06:21 AM

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