MHB Defining Branch Cuts for f(z) = z^(1/3)

[Dustinsfl]

Given the function \(f(z) = z^{1/3}\). Can the branch cuts be defined by the rays from zero through the three roots of unity and to infinity?

 

[mathbalarka]

Branch cuts are curves on $\hat{\Bbb C} = \Bbb C \cup \{\infty\}$ removal of which gives a domain $D \subset \hat{\Bbb C}$ over which $f(z)$ is holomorphic.

Can the branch cuts be defined by the rays from zero through the three roots of unity and to infinity?

The sheets of $f(z) = z^{1/3}$ are branched only at $z = 0$ in the complex plane. Why the three roots of unities? The only branches of $z^{1/3}$ on $\Bbb P^1$ are $z = 0$ and $z = \infty$ so take any closed curve $\Gamma$ joining those points on the Riemann sphere.

 

[Dustinsfl]

Branch cuts are curves on $\hat{\Bbb C} = \Bbb C \cup \{\infty\}$ removal of which gives a domain $D \subset \hat{\Bbb C}$ over which $f(z)$ is holomorphic.
The sheets of $f(z) = z^{1/3}$ are branched only at $z = 0$ in the complex plane. Why the three roots of unities? The only branches of $z^{1/3}$ on $\Bbb P^1$ are $z = 0$ and $z = \infty$ so take any closed curve $\Gamma$ joining those points on the Riemann sphere.

There should be three branch cuts though from 0 out to infinity. Is it not common to define it through the rays that would connect to the roots of unity?

 

[mathbalarka]

Why the roots of unities? Branch cuts are usually taken through *branch points* of $z^{1/3}$. What are the branch points of $w = z^{1/3}$ over $\hat{\Bbb C}$?

There should be three branch cuts though from 0 out to infinity.

Are you implying that you want to draw three branch cuts from $0$ to $\infty$, $\omega \infty$ and $\omega^2 \infty$ respectively? Please clarify.

 

[Dustinsfl]

Why the roots of unities? Branch cuts are usually taken through *branch points* of $z^{1/3}$. What are the branch points of $w = z^{1/3}$ over $\hat{\Bbb C}$?
Are you implying that you want to draw three branch cuts from $0$ to $\infty$, $\omega \infty$ and $\omega^2 \infty$ respectively? Please clarify.

That is what I said in the first post.

 

[mathbalarka]

That is what I said in the first post.

It's unnecessary in that case. $\infty, \omega \infty, \omega^2 \infty$ all identifies to the point at infinity in the Riemann sphere $\Bbb P^1$. A single slit through $0$ to $\infty$ is sufficient.

 

[Dustinsfl]

It's unnecessary in that case. $\infty, \omega \infty, \omega^2 \infty$ all identifies to the point at infinity in the Riemann sphere $\Bbb P^1$. A single slit through $0$ to $\infty$ is sufficient.

What are the other two branch cuts then? Functions of the form \(z^{m/k}\) where \(gcd(m,k) = 1\) have \(k\) branch cuts. If you are saying the other two branch cuts are the same, where are the other two at then?

 

[mathbalarka]

Functions of the form \(z^{m/k}\) where \(gcd(m,k) = 1\) have \(k\) branch cuts.

Where did you find that? It's false : $z^{1/n}$ has a single branch cut from $0$ to $-\infty$.

Perhaps whatever text included that meant that there must be $n$ branch cuts on *each* sheet (# of sheets are $n$) of the Riemann surface?

 

[Dustinsfl]

Where did you find that? It's false : $z^{1/n}$ has a single branch cut from $0$ to $-\infty$.

Perhaps whatever text included that meant that there must be $n$ branch cuts on *each* sheet (# of sheets are $n$) of the Riemann surface?

The branch cuts don't occur where the sheets of Riemann surface are split then?

 

[mathbalarka]

The branch cuts don't occur where the sheets of Riemann surface are split then?

Not sure what you mean. Do you know what a cut is? It appears on *the complex plane* NOT on the Riemann surface. The sheets are just copies of $\Bbb C$ along which they are cut and identified properly.