My Math Forum  

Go Back   My Math Forum > College Math Forum > Complex Analysis

Complex Analysis Complex Analysis Math Forum


Thanks Tree4Thanks
  • 2 Post By romsek
  • 1 Post By romsek
  • 1 Post By romsek
Reply
 
LinkBack Thread Tools Display Modes
November 2nd, 2017, 03:29 AM   #1
Newbie
 
Joined: Nov 2017
From: Portugal

Posts: 6
Thanks: 0

Post 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?
ghostwin is offline  
 
November 2nd, 2017, 04:02 AM   #2
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: Southern California, USA

Posts: 1,604
Thanks: 817

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 $z_0=1$ ?
romsek is offline  
November 2nd, 2017, 01:21 PM   #3
Newbie
 
Joined: Nov 2017
From: Portugal

Posts: 6
Thanks: 0

Reply

Yeh, im sorry. I mean what you just said.
ghostwin is offline  
November 2nd, 2017, 01:52 PM   #4
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: Southern California, USA

Posts: 1,604
Thanks: 817

$z = (z-1) + 1$

$\dfrac{i}{z+1} =\dfrac{i}{(z-1)+2} = \dfrac i 2 \dfrac{1}{1+\frac{z-1}{2}}$

$\dfrac{1}{1+\frac z a} = \sum \limits_{k=0}^\infty~(-1)^k \left(\dfrac z a\right)^k$

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
romsek is offline  
November 2nd, 2017, 02:24 PM   #5
Newbie
 
Joined: Nov 2017
From: Portugal

Posts: 6
Thanks: 0

Reply

Thank you! And it possible to compute when z_{0} = -1?
ghostwin is offline  
November 2nd, 2017, 03:09 PM   #6
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: Southern California, USA

Posts: 1,604
Thanks: 817

Quote:
Originally Posted by ghostwin View Post
Thank you! And it possible to compute when z_{0} = -1?
just apply the same method
Thanks from ghostwin
romsek is offline  
November 2nd, 2017, 03:41 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?
ghostwin is offline  
November 2nd, 2017, 03:50 PM   #8
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: Southern California, USA

Posts: 1,604
Thanks: 817

Quote:
Originally Posted by romsek View Post
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

$c_{-1} = i$

$c_0 = -1$

$c_1 = 1$
Thanks from ghostwin
romsek is offline  
November 2nd, 2017, 03:56 PM   #9
Newbie
 
Joined: Nov 2017
From: Portugal

Posts: 6
Thanks: 0

Reply

Sorry for all my questions! But thank you so much!!
ghostwin is offline  
Reply

  My Math Forum > College Math Forum > Complex Analysis

Tags
i or z, laurent, series



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Laurent series Robotboyx9 Calculus 2 October 25th, 2017 09:07 AM
Laurent Series? yusuf zeyd Complex Analysis 0 May 17th, 2015 03:37 AM
Is finding laurent series expansion of f at z_0 using geometric series convenient? king.oslo Complex Analysis 0 December 28th, 2014 07:50 AM
Laurent Series WWRtelescoping Complex Analysis 7 May 6th, 2014 12:09 AM
laurent series and the new one capea Complex Analysis 1 October 25th, 2011 07:21 AM





Copyright © 2017 My Math Forum. All rights reserved.