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Homework Help: Complex Contour Integral 2

  1. May 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Evaluate each of the following by Cauchy's Integral formula

    a)## \int_cj \frac{\cos z}{3z-3\pi} dz## c1: |z|=3, c2:|z|=4

    b) ##\int_c \frac{e^{3z}}{z-ln(2)} dz## c=square with corners at ##\pm(1\pm i)##



    2. Relevant equations
    ##f(z_0)=\frac{1}{2 \pi i}\int_c \frac{f(z)}{z-z_0}##


    3. The attempt at a solution

    a) ## \int_{c_j} \frac{\cos z}{3z-3\pi} dz## c1: |z|=3, c2:|z|=4

    ##\frac{1}{3}\int_{c1} \frac{\cos z}{z- \pi} dz =0## since ##\pi## lies outside c1 and hence ##\frac{\cos z}{z- \pi}## is analytic on and inside c1

    ##\frac{1}{3}\int_{c1} \frac{\cos z}{z- \pi} dz = \frac{1}{3} (2\pi i) \cos (\pi)= -\frac{2}{3} \pi i## since ##\pi## lies inside c2




    b) ##\int_c \frac{e^{3z}}{z-ln(2)} dz## c=square with corners at ##\pm(1\pm i)##

    ##\int_c \frac{e^{3z}}{z-ln(2)} dz=2 \pi i e^{3 ln(2)}=2^4 \pi i##...?

    Thanks
     
  2. jcsd
  3. May 1, 2012 #2
    Looks ok to me.
     
  4. May 2, 2012 #3
    In b) if we had ##\int_c \frac{e^{3z}}{(z-ln(2))^3} dz##

    We'd get the same answer because we have ##(z-ln(2))^3=0 \implies z-ln(2)=0##..?
     
  5. May 2, 2012 #4
    No.
    [tex]f^{(n)}(a)=\frac{n!}{2\pi i}\oint \frac{f(z)}{(z-a)^{n+1}}dz[/tex]
     
  6. May 3, 2012 #5
    ## \displaystyle \oint \frac{e^{3z}}{(z-ln(2))^{3}}dz=\frac{2\pi i}{3!} (3e^{3(ln 2)})=2^4 \pi i##..which is the same as original q part b)?
     
  7. May 3, 2012 #6
    When I plug your integral into that formula, I get:

    [tex]\frac{d^2}{dz^2}\left(e^{3z}\right)\biggr|_{z=\ln(2)}=\frac{2!}{2\pi i} \oint \frac{e^{3z}}{(z-\ln(2))^3}dz[/tex]

    or

    [tex]\oint \frac{e^{3z}}{(z-\ln(2))^3}dz=72\pi i[/tex]

    Also, try and learn to check them in Mathematica:

    Code (Text):

    NIntegrate[(Exp[3*z]/(z - Log[2])^3)*2*I*
        Exp[I*t] /. z -> 2*Exp[I*t], {t, 0, 2*Pi}]
    N[72*Pi*I]


    -1.4210854715202004*^-14 + 226.19467105847158*I

    0. + 226.1946710584651*I
     
     
  8. May 3, 2012 #7
    I didnt differentiate twice!

    Thank you!
     
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