# branch cuts

• Mar 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.
• Mar 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 $\displaystyle \log z$. This function is defined for all $\displaystyle z\not = 0$ as $\displaystyle \log z = \ln |z| + i \arg (z)$ where $\displaystyle \ln$ is the regular natural logarithm and $\displaystyle \arg$ is the argument on $\displaystyle (-\pi,\pi]$. Now, this function is continous everywhere except on the line segment $\displaystyle (-\infty,0]$. It can be shown that $\displaystyle \log z$ is analytic everywhere except on $\displaystyle (-\infty,0]$ which we call "its branch". Note, if we defined the argument function on the interval $\displaystyle [0,2\pi)$ then our newly defined logarithm will have branch $\displaystyle [0,\infty)$ but the former is the one usually used. Thus, when the problem asks to find the branch of $\displaystyle \log (z^3 - 2)$ think of it as asking "where is the function non-analytic". From above, that happens when $\displaystyle z^3 - 2$ is on the line segment $\displaystyle (-\infty,0]$ and that happens when $\displaystyle z\in (-\infty, \sqrt[3]{2}]$.
• Mar 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.
• Mar 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 $\displaystyle \arg_1 z$ is defined on branch $\displaystyle [0,2\pi)$ and $\displaystyle \arg_2 z$ is defined on branch $\displaystyle (-\pi,\pi]$. Now consider the argument of $\displaystyle -i$ and see what you get.