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Cauchy-Riemann equations

  1. Aug 4, 2012 #1
    1. The problem statement, all variables and given/known data

    write down the C-R equations and use them to determine those points at which the following functions are analytic
    (i)h(z)=[itex]x^2-y^2-x + i(2xy+y) [/itex]
    (ii)h(z)=cos2xcosh2y - isin2xsinh2y

    2. Relevant equations



    3. The attempt at a solution

    ok so C-R equations are for z=u+iv eq1 = [itex] \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} [/itex] and eq2= [itex] \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} [/itex]

    for for (i) i get [itex]\frac{\partial u}{\partial x} = 2x-1 and \frac{\partial v}{\partial y} = 2x+1 [/itex]
    and [itex]\frac{\partial v}{\partial x} = 2y and \frac{\partial u}{\partial y} = -2y [/itex] equation 2 holds but equation 1 does not hold. So am i right in thinking the function is not analytic? both equations have to be satisfied right? its the way the question is worded is throwing me "those points" ...
     
  2. jcsd
  3. Aug 4, 2012 #2

    HallsofIvy

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    A function is "analytic" at a given point if and only if it satisfies the C-R equation in some neighborhood of that point. Here, you are correct that the first function is not analytic for any points and the second function is analytic for all points.
     
  4. Aug 4, 2012 #3
    and for part (ii) i got the reverse
    as in [itex] \frac{\partial u}{\partial x} = -\frac{\partial v}{\partial y} [/itex]
    [itex] \frac{\partial u}{\partial y} = \frac{\partial v}{\partial x} [/itex]

    which leads me to believe this function isnt analytic either...
     
  5. Aug 4, 2012 #4
    Hi Hallsofivy,
    Thanks a million for that. I didnt realise youd got in there between my question and me reply!
    So I guess what your saying is it doesnt matter which equation has the negative?
     
  6. Aug 6, 2012 #5
    A function is analytic if (slightly hand-wavingly) it only depends on z, and not it's complex conjugate. You can see that in i), your function can be written as [itex] f(z) = z^2 - \bar{z} [/itex], meaning it's not analytic for sure.

    If you have the C-R-equations holding with an additional minus sign, then the function only depends on complex conjugate of z (and not z itself). I think people call this kind of functions antianalytic.
     
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