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Laurent series

  1. Nov 29, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-11-29_13-28-43.png
    and the solution (just to check my work)
    upload_2016-11-29_13-36-0.png
    2. Relevant equations
    None specifically. There seems to be many ways to solve these problems, but the one used in class seemed to be partial fractions and Taylor series.

    3. The attempt at a solution
    The first step seems to be expanding this using partial fractions, giving me
    upload_2016-11-29_13-32-27.png
    Now, for 0 < |z| < 1, we expand each of the fractions in the parenthesis in powers of z.
    upload_2016-11-29_13-34-4.png
    This is the Laurent series for f (z) which is valid in the region 0 < |z| < 1. I then need to get the other two series, which the next one I should try to get is for the region |z| > 2. To get that, it is suggested that I write the two partial fractions as:
    upload_2016-11-29_13-41-23.png
    However I am not sure what to do with this. I have seen things saying I should expand these two functions, and then add them together, however this does not give me the answer for the region |z| > 2, (in fact, it just gives me the first series, but a degree higher, which makes sense).
     
  2. jcsd
  3. Nov 29, 2016 #2

    vela

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    The reason for writing
    $$-\frac{1}{z-1} = -\frac 1z\left(\frac{1}{1-\frac 1z}\right)$$ and then expanding is because when ##|z|>1##, you have ##|1/z| < 1## so a series in positive powers of (1/z) will converge. If you look at the way you expanded 1/(z-2), you should see that you'll get a series of positive powers of (z/2), which won't converge for |z|>2.
     
  4. Nov 29, 2016 #3
    Thanks! Right, I okay, I see what I did wrong there, and I was able to get the correct series solution for |z| > 2.

    The only question I have left is how to get the 1< |z| < 2 series. I am not sure where to start there.
     
  5. Dec 7, 2016 #4
    $$\frac{1}{z}\left(\frac{1}{z-2} - \frac{1}{z-1}\right)= \frac{1}{z}\left(\frac{1}{-2}\cdot\frac{1}{1-\frac{z}{2}} - \frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}}\right)$$
    For ##1<|z|<2##, we have, ##\left|\frac{1}{z}\right|<1## and ##\left|\frac{z}{2}\right|<1##. So you can now use the formula for infinite geometric series.
     
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