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Expansion of complex function

  1. Jun 13, 2012 #1
    I have an exercise that says the following:

    Expand the following function as a series:

    f(z) = [itex]\frac{1}{z-1}[/itex] + [itex]\frac{1}{2-z}[/itex] for 1<lzl<2

    The result is attached, but I don't really understand what has been done. Therefore tell me:
    How is that series generated? Initially I thought I should do a taylor expansion but then that is always expanded about a specific point, which is not given in the exercise.
     

    Attached Files:

  2. jcsd
  3. Jun 13, 2012 #2

    vela

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    You're expanding about z=0. You can tell from the region for which you're asked to find the series. It's an annulus centered at z=0.

    Look up Laurent series.
     
  4. Jun 13, 2012 #3
    okay, I have done and I think I understand it mostly. But I still don't understand how you see that one MUST expand about z=0 - isn't that just done because it is the easiest?
     
  5. Jun 13, 2012 #4
    Hmm.. Now I've been looking a lot on this exercise, and you need to help me. I do understand that a function which is not analytic at certain singularities can be expanded as a laurent series about them. However, in this exercise f is only defined in the region 1<lzl<2 and is thus analytic everywhere. Therefore, why would a taylor series about a point z0 which satisfies 1<lz0l<2 not work?
    I don't really understand what is done to achieve the attached result. The solution notes says that the first expression (which I guess must be 1/(z-1)) is written as a series for z>1 and the second expression (which I again must guess refers to 1/(2-z)) is expanded for z<2. But what on earth is this supposed to mean? If the functions are expanded around z=0 the above can't be true, and either way you get terms for (1/z)k as though a laurent series is written.

    Can you please show how to do it, and I will learn from that rather than having to guess my way out of this (my book's chapter on Laurent series is VERY poor).
     
  6. Jun 13, 2012 #5
    1 < |z-0|< 2

    If it were centered anywhere else, the 0 would be replaced by that number. Remember that |z| < 2 => z in D(0;2)

    (I'm on my phone so Latex is a little time consuming to do).
     
  7. Jun 13, 2012 #6

    vela

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    No. Such a series won't converge for all points in the annulus. Suppose we expanded 1/(1-z) about z=3/2, for example. First, do some algebra to get
    $$\frac{1}{1-z} = \frac{1}{1-(z-3/2)-3/2} = -\frac{1}{1/2} \left(\frac{1}{1+2(z-3/2)}\right).$$ Then use the fact that
    $$\frac{1}{1+w} = 1-w+w^2-w^3+\cdots$$ with w=2(z-3/2) to obtain the series
    $$\frac{1}{1-z} = -2\left[1-2(z-3/2)+4(z-3/2)^2-8(z-3/2)^3+\cdots\right] = -2+4(z-3/2)-8(z-3/2)^2+\cdots.$$ (You could also find the Taylor series by taking derivatives, but this method is much faster.) This series will converge only when |w|<1, which means that |z-3/2| < 1/2. In other words, it'll only converge for points in the complex plane contained in a circle of radius 1/2 centered at z=3/2. Most of points in the annulus 1<|z|<2 don't meet this criterion, e.g. z=-3/2.

    Consider the Taylor series for 1/(1-z).
    $$\frac{1}{1-z} = 1+z+z^2+z^3+\cdots.$$ This series is valid if |z|<1; it won't converge if |z|>1. So for |z|>1, you expand in powers of 1/z. That series will converge because |1/z| will be less than 1.
    $$\frac{1}{1-z} = \frac{1}{z(1/z-1)} = -\frac{1}{z}\left(\frac{1}{1-1/z}\right) = -\frac{1}{z}\sum_{n=0}^\infty \left(\frac{1}{z}\right)^n.$$ For your original function, the two singularities divide the complex plane into three regions: |z|<1, 1<|z|<2, and |z|>2. For each region, you expand each term in either powers of z or powers of 1/z so that they'll both converge.

     
    Last edited: Jun 13, 2012
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