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Math Help - 2 more questions in Residue Theorem-Complex

  1. #1
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    2 more questions in Residue Theorem-Complex

    1. How can I calculate the residue in z=1 of the next function:
     f(z)=(2z-1)e^{\frac{z}{z-1}} ??
    I've tried using Laurent expansion but couldn't figure out how to expand this function into a series (using the known expansion of the exponent don't help us here... ).

    2. Calculate the next Integral:
     \frac{1}{2\pi i} \int_{|z|=1} \frac{z^{n-1}}{3z^n-1} dz .

    My problem with this question is that we have n singularities and they are all inside our contour... I've tried to paramterize but without any success..
    Hope you'll be able to help me



    Thanks in advance!
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  2. #2
    Junior Member
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    I believe the answer to the second part of your question is 1/3....

    (1/2 \pi i) \oint\frac{1/3z^{n-1}dz}{z^n-1/3} = \sum_{k=0}^{n-1}\frac{1/3(\zeta _{k})^{(n-1)/n}}{n(1/3)^{(n-1)/n}}e^{2\pi ik(n-1)/n} with \zeta _k = (1/3)^{1/n}e^{2\pi ik/n} for k=0,1,2.....,n-1

    Basically just sum over the residues evaluated at the n roots of 1/3...

    Hope this helps...
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  3. #3
    Super Member
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    Quote Originally Posted by WannaBe View Post
    1. How can I calculate the residue in z=1 of the next function:
     f(z)=(2z-1)e^{\frac{z}{z-1}}
    (2z-1)e^{z/(z-1)}=(2z-1)e\sum_{n=0}^{\infty}\frac{1}{n!(z-1)^n}

    =2e\sum_{n=0}^{\infty}\frac{(z-1)^{1-n}}{n!}+e\sum_{n=0}^{\infty}\frac{1}{n!(z-1)^n}

    and the coefficient on the 1/(z-1) term is:

    2e(1/2)+e=2e

    which is consistent with the numerical result:

    Code:
    In[13]:=
    NIntegrate[(2*z - 1)*Exp[z/(z - 1)]*(1/4)*
        I*Exp[I*t] /. 
       z -> 1 + (1/4)*Exp[I*t], {t, 0, 2*Pi}]
    N[4*Pi*I*E]
    
    Out[13]=
    7.105427357601002*^-15 + 
      34.15893689068571*I
    
    Out[14]=
    0. + 34.158936890694264*I
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