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Compute the taylor's expansion (series)

  1. May 9, 2008 #1
    [SOLVED] compute the taylor's expansion (series)

    1. The problem statement, all variables and given/known data

    compute the taylor's expansion (series) for:

    f(z)= [z^4 + (2-3i)*z^3 - 6i*z^2 + 2]/[z(z+2)]

    where Zo (Z node or Z(0)) = 1



    i need answer as soon as possible please!
     
  2. jcsd
  3. May 9, 2008 #2

    malawi_glenn

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    we don't give answer here, read and follow the forum rules.
     
  4. May 9, 2008 #3
    i tried replacing each z wth x+iy but it went into a huge binomial equation i couldnt solve, here is part of my work:
    z^4 = (x+iy)^4 = x^4 + (x^3 + iy) + (x^2 - y^2) + (x - iy^3) + y^4

    z^4 + (2-3i)*z^3 = x^4 + (3-3i)*(x^3 + x^2 + x - y^2) + (3 + 3i)*(y - y^3) + y^4

    and i dont know what to do after that, please help me!
    thanks...
     
  5. May 9, 2008 #4

    Vid

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    Start from the definition of a Taylor series.
     
  6. May 9, 2008 #5

    malawi_glenn

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    this is complex analysis right?

    Are you supposed do evaluate taylor series for this f(z) around z_0 = 1 ?
     
  7. May 9, 2008 #6
    yes, can you help in that?
     
  8. May 9, 2008 #7
    the problem is that this equation will not be easy to defferenciate it has to be simplified first
    or if you could do it as it is please tell me how

    thanks
     
  9. May 9, 2008 #8

    Vid

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    Divide z+2 into the top. Break apart the resulting fraction and simplify.
     
  10. May 9, 2008 #9
    can you please show me the steps of dividing by z+2 & the breaking apart of the fraction?

    thanks
     
  11. May 9, 2008 #10

    Vid

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    No..
     
  12. May 9, 2008 #11

    malawi_glenn

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    We dont provide that kind of help here. You show that you can try first, then we help you.

    Takning the derivatives of [z^4 + (2-3i)*z^3 - 6i*z^2 + 2]/[z(z+2)] is straightforward if you know your derivation rules, which are the same as in real analysis.
     
  13. May 9, 2008 #12
    ok thanks very much, i'll try solving it again
     
  14. May 11, 2008 #13
    thanks for all your help, i've finally done it, and here is my final answer:

    f(z)= (5/3 - 3i) + (10/9 - 3i) (z - 1)/1! + (50/27) [(z-1)^2]/2! + (4/27) [(z - 1)^3]/3! + ....
     
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