Complex Analysis Question

[spacenerd]

I was hoping someone could clarify the idea of a branch cut for me. In class, my professor talked about how a branch cut is used to remove discontinuities. He gave an example of |z|=1 needing a branch cut along the positive real axis. If this because going from 0 to 2\pi, the 0 and 2\pi match up?

 

[lanedance]

|z|=1 is not a function

 

[Char. Limit]

It seems like an implicitly defined function to me.

 

[lanedance]

how do you figure?

|z|=1 seems to me like maybe an example of a path you would take in C, moving around the unit circle to show a given function is multivalued

 

[spacenerd]

Right, Well I guess I should have specified that z was an element of the Complex plane. Thoght it was kinda implied by the post title.
How about ln(z). I know that this is defined as;
ln(z)=ln|z|+i*arg(z), where 0<arg(z)<2pi.
This requires a branch cut to not include 0 and 2pi.
I'm just a little fuzzy on the notion of a branch cut.

 

[lanedance]

it was clear z was an element of the complex plane, as per normal notation

what wasn't clear was which function you were working with. |z|=1 is not a function, it is a constraint, which represents the set of points on the unit circle in the complex plane.

when a function is multi-valued, you can choose where to put branch cuts in the complex plane so that no path that does not cross the branch cut is able to take you to a multivalued, ie for any path that crosses the same point z $f(z) = f(z)$ always

the location of a branch cut is not in general unique,you can chose where to upt it, however often it is specified by certain points