Finding Maclaurin & Laurent Series for f(z)

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SUMMARY

The discussion focuses on finding the Maclaurin and Laurent series for the function f(z) = (z + 2)/(z - 2). The Maclaurin series is derived for the domain |z| < 2, while the Laurent series is centered at z0 = 0 for the domain 2 < |z| < ∞. The transformation of the function into a suitable form for series expansion is emphasized, particularly utilizing the geometric series approach.

PREREQUISITES
  • Understanding of Maclaurin series expansion
  • Familiarity with Laurent series and their applications
  • Knowledge of geometric series and their convergence
  • Basic complex analysis concepts
NEXT STEPS
  • Study the derivation of Maclaurin series for various functions
  • Learn about the conditions for convergence of Laurent series
  • Explore geometric series and their applications in complex functions
  • Investigate the implications of singularities in complex analysis
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Students and educators in mathematics, particularly those studying complex analysis, as well as anyone interested in series expansions and their applications in theoretical and applied mathematics.

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Homework Statement



f(z) = (z + 2)/(z - 2)

a) Find the Maclaurin Series for f on the doman |z| < 2.

b) Find the Laurent Series for f centered at z0 = 0 on domain 2 < |Z| < inf.


Homework Equations




The Attempt at a Solution



I'm having a hard time figuring out how (z + 2)/(z - 2) = 1 + (4/(z-2)) = 1 - (2/(1 - (z/2)).

I tried referring to a geometric series, but I don't think I have the right approach.
 
Last edited:
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hmmmm... been a while but how about
[tex]\frac{z+2}{z-2} = (\frac{1/z}{1/z})\frac{z+2}{z-2} = \frac{1+2/z}{1-2/z} = (1+\frac{2}{z})\frac{1}{1-2/z}[/tex]
 
or for your specific question working back
[tex]1 + \frac{4}{z-2} = \frac{z-2}{z-2} + \frac{4}{z-2} = \frac{z-2+4}{z-2} = \frac{z+2}{z-2}[/tex]
is that your question?
 

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