text: # Laurent series
  
  1. Apr 10, 2009
  
  ### krindik
  
  Hi,
  can u pls help me on this?
  
  1. The problem statement, all variables and given/known data
  Find Laurent series that converges for
  
  $$\, 0 < |z - z_0| < R }$$ and determine precise region of convergance
  
  $$\, \frac {1}{z^2 + 1} \,\,$$
  
  2. Relevant equations
  
  3. The attempt at a solution
  I tried to spilt this into fractions
  i.e
  $$f(x) \, = \, \frac{A}{z-i} + \, \frac{B}{z+i}$$
  
  as I would have done for
  
  $$\frac {1}{z^2 - 1} \,\,$$
  
  But in that case I would expand it with a geometrical series.
  The problem rises with $$i$$ instead of $$1$$
  
  2. The problem statement, all variables and given/known data
  Can u pls explain how can I choose the method of expansion (Laurent, Taylor) given a function f(x) ?
  
  Thanks
  1. The problem statement, all variables and given/known data
  
  2. Relevant equations
  
  3. The attempt at a solution
  
  2. Apr 10, 2009
  
  ### latentcorpse
  
  can't you just say $w=z^2$ then you can make
  
  $\frac{1}{w-1}=\frac{1}{w} \frac{1}{1-\frac{1}{w}}$ and expand as a geometric series...
  
  3. Apr 10, 2009
  
  ### krindik
  
  Thanks.
  
  In fact the question is
  $$\, \frac {1}{z^2 + 1} \,\,$$
  
  not
  
  $$\, \frac {1}{z^2 - 1} \,\,$$
  
  I'll restate the question
  
  Find Laurent series that converges for
  
  $$\, 0 < |z - z_0| < R }$$
  
  and determine precise region of convergance for
  $$\, \frac {1}{z^2 + 1} \,\,$$
  
  where
  $$z_0 = i$$
  
  Additionally, I would like to know how u use $$\, 0 < |z - z_0| < R }$$
  in coming up with a solution
  
  Last edited: Apr 10, 2009
  4. Apr 10, 2009
  
  ### Dick
  
  The Laurent series will be an expansion in powers of (z-i). Use your partial fractions form. The A/(z-i) is the (-1) power. Write the B/(z+i) part as B/(2i+(z-i)) and do the geometric series expansion with (z-i) as the variable.
  
  5. Apr 10, 2009
  
  ### latentcorpse
  
  why can we not just change variable here?
  
  6. Apr 10, 2009
  
  ### Dick
  
  You can. Isn't that basically what I did when I wrote z+i=(z-i)+2i?
  
  7. Apr 10, 2009
  
  ### krindik
  
  Thank you,
  
  I have few questions
  1. Is it that you choose $(z - i)$ terms because the question specifies it should converge in $0 < |z-i| < R$ ?
  
  2. So $\frac{B}{z + i}$ will be expanded like
  
  $\frac{1}{2i} * \frac{1}{1+\frac{z-i}{2i}} = \sum_{n=0}^{\infty}{s^n}$
  
  where
  $s = \frac{z-i}{2i}$
  
  so the answer to the solution would be
  $\frac{A}{z-i} + \{expansion\, in\, 2\}$ ?
  
  Thanks again
  
  8. Apr 11, 2009
  
  ### Dick
  
  I chose (z-i) because you specified that you want to expand around z0=i in post 3. I couldn't tell where you wanted to put the expansion from the first post. And yes, you've got the expansion right.
  
  9. Apr 11, 2009
  
  ### krindik
  
  Thank you very much Dick
  
  I shall summarize here, so that anybody else following this thread can get the point.
  
  $\frac{1}{1+z^2} = \frac{A}{z-i} + \frac{B}{z+i}$
  
  expand
  $\frac{B}{z+i} = \frac{B}{2i} * \frac{1}{1-(-1 * \frac{z-i}{2i})}$
  
  then by expanding in geometric series
  $\frac{1}{1-(-1 * \frac{z-i}{2i})} = \sum_{n=0}^\infty{[(\frac{-1}{2i})^n * (z-i)^n]}$
  
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