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Math Help - Laurent series (Complex)

  1. #1
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    Laurent series (Complex)

    Using term-by-term differentiation, expand f(z)=\frac{1}{(z+3)^2} in a Laurent series valid for |z| > 3.

    Ok, so, suppose I get a few derivatives:

    f'(z)=-2(z+3)^{-3}

    f''(z)=6(z+3)^{-4}

    f'''(z)=-24(z+3)^{-5}
    ...

    There's infinitely many. What next?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by scorpion007 View Post
    Using term-by-term differentiation, expand f(z)=\frac{1}{(z+3)^2} in a Laurent series valid for |z| > 3.

    Ok, so, suppose I get a few derivatives:

    f'(z)=-2(z+3)^{-3}

    f''(z)=6(z+3)^{-4}

    f'''(z)=-24(z+3)^{-5}
    ...

    There's infinitely many. What next?
    Pattern.

    Do you have to do it this way? Note that f(z)=\frac{1}{z^2}\frac{1}{(1+\frac{3}{z})^2}
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  3. #3
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    I guess I have to, since that's what the question asks.

    I know the pattern is:

    f^{(n)}(z)=(n+1)!(-1)^n(z+3)^{-(n+2)}

    but I don't know what to do with it.
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  4. #4
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    Doing it the other way, I suppose what happens is:

    <br />
f(z)=\frac{1}{z^2}\frac{1}{(1+\frac{3}{z})^2} = \frac{1}{z^2}\left(\sum_{n=0}^\infty \left(-\frac{3}{z}\right)^n\right)^2<br />
for |z|>3.

    But the whole series is squared. How am I suppose to write out the first few terms?
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by scorpion007 View Post
    Doing it the other way, I suppose what happens is:

    <br />
f(z)=\frac{1}{z^2}\frac{1}{(1+\frac{3}{z})^2} = \frac{1}{z^2}\left(\sum_{n=0}^\infty \left(-\frac{3}{z}\right)^n\right)^2<br />
for |z|>3.

    But the whole series is squared. How am I suppose to write out the first few terms?
    Hint: \frac{1}{(1+z)^2}=-\left(\frac{1}{1+z}\right)'...
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  6. #6
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    Ok, so do I subs that in one of the factors and write it in terms of d/dz ?

    I'm not sure what do with that...
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by scorpion007 View Post
    Ok, so do I subs that in one of the factors and write it in terms of d/dz ?

    I'm not sure what do with that...
    \left(\frac{1}{1-x}\right)'=\left(\sum_{n\in\mathbb{N}}x^n\right)'=  \sum_{n\in\mathbb{N}}nx^{n-1}

    Adapt that.
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  8. #8
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    I guess you want me to take the deriv of both sides?

    f'(z) =\frac{2}{z^2}\sum_{n=0}^\infty \left(-\frac{3}{z}\right)^n\cdot \sum_{n=0}^\infty n \left(-\frac{3}{z}\right)^{n-1}\cdot \frac{3}{z^2}

    is that right? I could simplify this and all, but I don't quite see where this is going...
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  9. #9
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    Quote Originally Posted by Drexel28 View Post
    Hint: \frac{1}{(1+z)^2}=-\left(\frac{1}{1+z}\right)'...
    Maybe it's easier to use Newton's binomial theorem to expand (1+z)^{-2}.
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  10. #10
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    Ah, I figured it out guys.

    The example I needed was on this page: Pauls Online Notes : Calculus II - Power Series and Functions
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