branch cuts

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March 5th 2008, 08:16 PM
frankdent1

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.
March 5th 2008, 08:29 PM
ThePerfectHacker

Quote:

Originally Posted by frankdent1

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 $z\in (-\infty, \sqrt[3]{2}]$ .
March 5th 2008, 09:16 PM
frankdent1

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.
March 6th 2008, 12:52 PM
ThePerfectHacker

Quote:

Originally Posted by frankdent1

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.