∫ + γ z ¯ d z and ∫ + γ z d z, where the contour is z ( θ ) = 2 e i θ , 0 ≤ θ ≤ 2 πI thought these two integrals are equal to zero, but after calculation, the first equals to 8 π i and the second is equal to 2 ( e 4 π i − 1 ). Why is that? Are they, z ¯ and z, not continuous on C with some antiderivatives?

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∫          +      γ                  z      ¯        d  z and       ∫          +      γ        z  d  z, where the contour is   z  (  θ  )  =  2      e          i      θ      ,   0  ≤  θ  ≤  2  πI thought these two integrals are equal to zero, but after calculation, the first equals to   8  π  i and the second is equal to   2  (      e          4      π      i        −  1  ). Why is that? Are they,             z      ¯       and z, not continuous on       C   with some antiderivatives?

timmeraared · Accepted Answer

Step 1      ∫          γ        z  d  z  =  0 y Cauchy&#039;s theorem. You can also use the evident parameterization   z  =  2      e          i      θ      . Then, calculate       ∫    0          2      π        (      e          i      θ        )  (  2  i  ⋅  2      e          i      θ        )  d  θ  =  4  i      ∫    0          2      π            e          i      θ        d  θ  =  0.Step 2Using the same parameterization (here Cauchy&#039;s theorem does not apply), you get       ∫          γ            z    ¯    d  z  =      ∫    0          2      π        (      e          −      i      θ        )  (  2  i  ⋅  2      e          i      θ        )  d  θ  =  4  i      ∫    0          2      π        d  θ  =  8  π  i  .

excluderho · Accepted Answer

Step 1The other answer has already addressed the fact that your second integral is, in fact, zero. Another way to tackle the first integral is to recognize that along your given contour,   z      z    ¯    =      2    2      ⟹        z    ¯    =      4    z    ..Step 2Therefore, your first integral is just   8  π  i  ⋅      1          2      π      i            ∫          γ        1  ⋅            d      z              z      −      0        =  8  π  i by applying the Cauchy integral formula to the constant function 1.

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