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Question about Laurent Series

  1. Apr 21, 2007 #1
    Say we have the function:

    [tex]\frac{1}{\left( z-1 \right)\left( z+2 \right)}[/tex]

    Using partial fractions,

    [tex]\frac{1}{\left( z-1 \right)\left( z+2 \right)}\; =\; \frac{1}{z-2}\; -\; \frac{1}{z\; -\; 1}[/tex]

    My question comes in on why and how these equations are manipulted for different regions.

    Now for a) region |z| < 1

    [tex] \frac{1}{z-1}\; =\; -\frac{1}{1-z}\; =\; -\sum_{j=0}^{\infty }{z^{j}}\; [/tex]

    But for region 1 < |z| < 2

    [tex]\frac{1}{z-1}\; =\; \frac{1}{z}\frac{1}{1-\frac{1}{z}}\; =\; -\frac{1}{z}\sum_{j=0}^{\infty }{\frac{1}{z^{j}}\; =\; }\sum_{j=0}^{\infty }{\frac{1}{z^{j+1}}}[/tex]

    I have no idea how or why they are being manipulated for different regions. My book assumes me to be brilliant I suppose?
  2. jcsd
  3. Apr 21, 2007 #2
    No, it's not so hard.

    The key point is that the geometric series does not converge everywhere. So in each case, they are manipulating it so that the series converges in the desired region.
  4. Apr 21, 2007 #3
    Can you elaborate on how to make it converge for a region?
  5. Apr 22, 2007 #4

    Gib Z

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    Homework Helper

    Look up geometric series in google. For a series to converge, the terms should approach zero. That only happens when |z|<1 doesn't it?
  6. Apr 22, 2007 #5
    Sort of.

    [tex] \frac{1}{1-x}=\sum_{j=0}^{\infty }{x^{j}}[/tex]

    Is valid when |x| < 1. Thus in the region where |z| < 1, they just let x=z. But where |z| > 1, they let x=1/z in the geometric series formula above, so that the formula would make sense in the desired region.
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