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Math Help - Help with a Laurent expansion

  1. #1
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    Help with a Laurent expansion

    Hello. My book doesn't explain Laurent series expansions very well and so I was hoping for some help figuring out a problem from the exercises. The answers are in the book so I know I got the first two - I just don't understand the last two.

    'Find a Laurent series for the function 1/(z+z^2) in each of the following domains:

    a. 0 < |z| < 1 ...'

    For this one I did a partial fraction decomp and got 1/z - 1/(z+1).

    Finding a geometric series for the second term ... 1/(z+1) = 1/(1-(-z)) and since |-z| = |z| < 1, it has a geometric series representation \displaystyle\sum\limits_{j=0}^\infty (-z)^j.

    So, 1/(z+z^2) = 1/z - \displaystyle\sum\limits_{j=0}^\infty (-z)^j (So because z=0 is not defined in the annulus, 1/z is analytic - so it's OK to use and also has no series representation?) = \displaystyle\sum\limits_{j=-1}^\infty (-1)^{j+1}z^j.

    That and the domain (b) 1 < |z| I got OK - for (b) I factored out (1/z) from (1/z+1) to get a geometric series.

    My problem is when you move the annulus center to the other singularity...

    c. 0 < |z+1| < 1 { = - \displaystyle\sum\limits_{j=-1}^\infty (z+1)^j } and d. 1 < |z+1| --

    For c. I assume you can still use 1/z with no problems since the annulus doesn't includ z=0, but I am having a hard time trying to find a geometric series for 1/(z+1).

    Reverse triangle inequality didn't get me |z| < 1 (I got 0 < |z| < 2 - although I have doubts about that being correct); I don't think I can simply factor out 1/z to get 1/z*(1/(1+(1/z)) because it's possible for |z|<1 which would make |1/z|>1 ... so I just don't know what to try next.
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  2. #2
    MHF Contributor chisigma's Avatar
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    If You want to follow a very confortable way, then...

     \frac{1}{z+z^{2}} = \frac{1}{z}\cdot \frac{1}{1+z} = \frac{1}{z} -1 + z - z^{2} + \dots (1)

    It is easy to verify that (1) converges for 0 < |z| < 1...

    Kind regards

    \chi \sigma
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  3. #3
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    I was actually looking for help in the annulus 0 < |z+1| < 1, but I believe I found out how to do it. (Found a pretty good video on YouTube explaining Laurent series--)

    1/z+z^2 = 1/z * 1/(1+z)

    1/(1+z) stays as it is since z=-1 is the center of the "annulus" (punctured disk). (The 'why' I haven't completely figured out yet.)

    For 1/z = -1/(1-(1+z)) and we already know |1+z|<1, so that gives the geometric series - \displaystyle\sum\limits_{j=0}^\infty (1+z)^n .

    So we have - \displaystyle\sum\limits_{j=0}^\infty (1+z)^n * (1+z)^{-1} = - \displaystyle\sum\limits_{j=0}^\infty (1+z)^{n-1} = - \displaystyle\sum\limits_{j=-1}^\infty (1+z)^n. Hurray, case closed!
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