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Simple complex integration

  1. Aug 25, 2007 #1
    Integrate cosh(4z) w.r.t., z, for any path from [tex] \frac{-\pi i}{8} \inbox{to } \frac{\pi i}{8} [/tex]. If the function is analytic, i.e., obeys Cauchy - Riemann equations we can integrate as in standard calculus.
    [tex] \frac{{\partial \cosh(4(x+iy)}}{{\partial x}} = a\sinh(4x + 4iy) \\ \frac{{\partial \cosh(4x +4iy)}}{{\partial y }} = 4i\sinh(4x + 4iy)\\[/tex]. Therefore [tex] u_x [/tex] and [tex] v_y [/tex] are not equal, therefore the cosh(4z) is not analytic. So we integrate as follows: C (the path of integration) can be represented by z(t)= 0 + it, [tex] \frac{-\pi}{8} \leq t \leq \frac{\pi}{8}\\[/tex] Hence [tex] \dot z(t) = \iota \\[/tex] and f(z(t)) = cosh(it), therefore [tex] \int_C \cosh(z) dz = \int_{\frac{-\pi}{8}}^\frac{\pi}{8} \cosh(it)\iota dt \\[/tex] = sinh(it) evulated at the respective limits = [tex] 2\sinh(\frac{\iota\pi}{8}) \ [/tex]
    I don't think that this is correct, I think my Z(t) = 0 +it is wrong.
     
    Last edited: Aug 25, 2007
  2. jcsd
  3. Aug 25, 2007 #2

    Dick

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    cosh(4z) IS analytic. You have to split the function into real and imaginary parts before applying CR. cosh(z)=(e^z+e^-z)/2. Now e.g. e^z=e^(x+iy)=e^x*e^(iy)=e^x*(cos(y)+i*sin(y)). So the split for e^z is u(x,y)=e^x*cos(y) and v(x,y)=e^x*sin(y) and it IS analytic. Similarly for cosh(z).
     
  4. Aug 27, 2007 #3
    Thanks for the reply Dick. As a matter of interest do you know how "Cauchy" is pronounced? E.g., Ca-u-chy?
     
  5. Aug 27, 2007 #4

    Dick

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    Co-she, 'co' as in coordinate and 'she' as in she. You're welcome.
     
  6. Aug 27, 2007 #5
    You can use the fundamental theorem of calculus here, it is easy to find the primitive of cosh(4z)
     
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