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Integrating exponetial of z over the conjugate of z

  1. Sep 20, 2013 #1
    Im doing some complex variable "counter integration" problems and this one came up.

    [itex]I = \oint e ^{\frac{z}{\overline{z}}}dz[/itex]


    the integral must be done over a circle with radio r


    My first attempt was to do it in the exponetial form, so we have this:

    [itex]\frac{z}{\overline{z}} = e^{2i\theta} [/itex]

    but when i do so, i get an e to the power e to the power something and i start getting some long equation so i figure out, there must be an easier way to solve the problem. Hope someone can give me a hand. =D
    PS: sorry for my english, it´s my second language!
     
  2. jcsd
  3. Sep 20, 2013 #2

    Ray Vickson

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    All I can suggest is that you struggle with trying to do the integral! The final answer is very simple, but I cheated and used a computer algebra package to do it.
     
  4. Sep 21, 2013 #3
    Can you solve it with the Residue Theorem or Cauchy's Theorem? Well, how about this one:

    [tex]\mathop\oint\limits_{|u|=1} e^{u^2} du[/tex]

    How about this one then:

    [tex]\int_0^{2\pi} e^{e^{2it}} ri e^{it} dt[/tex]

    Then what happens if you just substitute [itex]z=re^{it}[/itex] into yours?
     
    Last edited: Sep 21, 2013
  5. Sep 21, 2013 #4
    Thanks for the answers, they really got me thinking and I just came up with an idea; now i do not know if its valid or logical in all its statements, but maybe you can tell me if it is..

    I can rewrite.
    [itex]e^{z/\overline{z}}[/itex]

    as

    [itex]e^{z^2/\left|z\right|^2}[/itex]

    then i get:

    [itex]\oint_{|z|=r} e^{z^2/\left|z\right|^2}dz[/itex]

    which is equal to
    [itex]\oint_{|z|=r} e^{z^2/r^2}dz[/itex]

    and when r=1 (in the unit circle) the integral is equal to zero(as it is an holomorfic/analytic function)

    [itex]\oint_{|z|=1} e^{z^2}dz = 0[/itex]

    then, as it is also holomorfic/analytic between |z|=1 and |z|=r the result must be the same because of the deformation theorm (its a rubber sheet geometry) so:

    [itex]\oint_{|z|=r} e^{z^2/\left|z\right|^2}dz=0[/itex]

    and finally:

    [itex]\oint_{|z|=r} e^{z/\overline{z}}dz=0[/itex]

    does it sound reasonable to you??
     
  6. Sep 21, 2013 #5
    Outstanding. Better than my way. But you could have stopped at just:

    [tex]\oint_{|z|=r} e^{z^2/r^2}dz=0, \quad r>0[/tex]

    since it's holomorphic. And also, not hard to numerically verify it in Mathematica in case you need to check something else that's not so obvious:

    Code (Text):

    In[1]:=
    NIntegrate[Exp[z/Conjugate[z]]*I*Exp[I*t] /.
       z -> Exp[I*t], {t, 0, 2*Pi}]

    During evaluation of In[1]:= NIntegrate::ncvb:NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in t near {t} = {1.1269*10^-7}. NIntegrate obtained -4.996*10^-16-1.27676*10^-15 I and 3.849598995114243`*^-12 for the integral and error estimates. >>

    Out[1]=
    -4.996003610813204*^-16 - 1.27675647831893*^-15*I
     
    That's basically zero.
     
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