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The Cauchy Riemann Equations for this Function

  1. Jun 26, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data

    Verify that each of the following functions is entire:
    f(z)=(z^2-2)e^(-x)e^(-iy)

    2. Relevant equations

    The Cauchy Riemann equations u(x,y) = ______ and v(x,y) = ______
    u_y=-v_x
    u_x=v_y

    3. The attempt at a solution

    So, I've done a few of these problems and understand that to prove it's entire we need to show that the C-R equations exist for all z and are satisfied. Also, noting that Ux Uy Vx Vy are continuous.

    My problem is, I'm not sure how to get this into correct form. What is u(x,y) =?
    Do I distribute first? How to change this function into something that works?

    Would it be
    (z^2-2)/(e^x*e^iy)
    Is that a good start? :/
     
  2. jcsd
  3. Jun 26, 2015 #2

    Mark44

    Staff: Mentor

    I don't believe so.

    For starters, write z = x + iy, and split ##(z^2 - 2)e^{-x}e^{-iy}## into its real and imaginary parts. u(x, y) is the real part of f(z) and v(x, y) is the imaginary part.

    A useful formula that you omitted, and that might come in handy, is ##e^{it} = \cos(t) + i\sin(t)##.
     
  4. Jun 26, 2015 #3

    RJLiberator

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    Thanks Mark.

    So we do have to expand the z then.

    So this would mean:

    (x^2+2yxi-y^2-2)/(e^xcos(y)+e^x(isin(y)))

    The thing that is throwing me off here is the negative exponents, I'm not sure what to do with those, so I put the term in the denominator.

    That doesn't seem right however.
     
  5. Jun 26, 2015 #4

    Mark44

    Staff: Mentor

    Moving them to the denominator is no help. Leave the e-x factor as is, and rewrite the e-iy factor as below.
    ##e^{-x}e^{-iy} = e^{-x}(\cos(y) - i\sin(y))##
     
  6. Jun 26, 2015 #5

    HallsofIvy

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    Rather than put "[itex]e^xcos(y)+e^x(isin(y))[/itex]" in the denominator, I think I would write [itex]e^{-x- iy}= e^{-x}e^{-iy}= e^{-x}(cos(-y)+ i sin(-y))= e^{-x}(cos(x)- i sin(x))[/itex]
     
  7. Jun 26, 2015 #6

    RJLiberator

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    That was indeed the element of this question that I was missing out/overlooked.

    So with that in play, we see that
    u(x,y) = (x^2-y^2-2)e^(-x)cos(x)+2ye^(-x)sin(y)
    v(x,y) = 2ye^(-x)cos(x)-(x^2-y^2-2)e^(-x)sin(y)
     
  8. Jun 26, 2015 #7

    RJLiberator

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    Re: this thread, I found out that the composition of two functions that are entire is entire.

    So One can look at (z^2-2) and e^(-x)e^(-iy) as two different functions, evaluate their cauchy-riemann equations and see that each is entire to conclude the composition is entire.

    =)
     
  9. Jun 26, 2015 #8

    WWGD

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    But notice you are not composing the functions, you are multiplying them ( though the result is true that the product of entires is entire )
     
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