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Math Help - branch cuts

  1. #1
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    branch cuts

    I'm having trouble understand branch cuts. I understand that a function is discontinuous along their branch cut but what I don't understand is how you can change the branch cut in order to solve equations.
    Example: determine a branch of f(z) = log(z^3-2) that is analytic at z=0 and find f(0) and f '(0).
    Here, the inside is -2 when z=0, so you'd choose a branch which isn't -Pi like -Pi/4. so f(0)=Log|-2| + Arg (-2). I guess my question is how do you find Arg(-2) with different branch cuts.
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  2. #2
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    Quote Originally Posted by frankdent1 View Post
    I'm having trouble understand branch cuts. I understand that a function is discontinuous along their branch cut but what I don't understand is how you can change the branch cut in order to solve equations.
    Example: determine a branch of f(z) = log(z^3-2) that is analytic at z=0 and find f(0) and f '(0).
    Here, the inside is -2 when z=0, so you'd choose a branch which isn't -Pi like -Pi/4. so f(0)=Log|-2| + Arg (-2). I guess my question is how do you find Arg(-2) with different branch cuts.
    This is how I like to think of branch cuts. Consider the function \log z. This function is defined for all z\not = 0 as \log z = \ln |z| + i \arg (z) where \ln is the regular natural logarithm and \arg is the argument on (-\pi,\pi]. Now, this function is continous everywhere except on the line segment (-\infty,0]. It can be shown that \log z is analytic everywhere except on (-\infty,0] which we call "its branch". Note, if we defined the argument function on the interval [0,2\pi) then our newly defined logarithm will have branch [0,\infty) but the former is the one usually used. Thus, when the problem asks to find the branch of \log (z^3 - 2) think of it as asking "where is the function non-analytic". From above, that happens when z^3 - 2 is on the line segment (-\infty,0] and that happens when z\in (-\infty, \sqrt[3]{2}].
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  3. #3
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    Say you find a branch where it's continuous at the point z=a + bi, does changing the branch from [-Pi,Pi) to [0,2Pi) or any other [theta, theta+2Pi) affect the value of Arg(z). I just see it as pointless to switch branches because there will always be a branch that will be continuous at that point, i think. Angle of z will always be the same no matter where you define your branch, i think.
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    Quote Originally Posted by frankdent1 View Post
    Say you find a branch where it's continuous at the point z=a + bi, does changing the branch from [-Pi,Pi) to [0,2Pi) or any other [theta, theta+2Pi) affect the value of Arg(z). I just see it as pointless to switch branches because there will always be a branch that will be continuous at that point, i think. Angle of z will always be the same no matter where you define your branch, i think.
    Not true. Say \arg_1 z is defined on branch [0,2\pi) and \arg_2 z is defined on branch (-\pi,\pi]. Now consider the argument of -i and see what you get.
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