MHB Is the Function f(z) = z^(1/3) Analytic?

[Dustinsfl]

I want to show \(f(z) = z^{1/3}\) is analytic. I need to show that for some domain D:
(1) the partial derivatives exist in D
(2) the C-R equations are satisfied
(3) f is continuous in D
(4) the partial derivatives are continuous in D
Then f is analytic in D.
\begin{alignat*}{3}
u_x
&= \frac{x\cos\Big(\frac{1}{3}\arctan\big(\frac{y}{x}\big)\Big) +
y\sin\Big(\frac{1}{3}\arctan\big(\frac{y}{x}\big)\Big)}
{3(x^2 + y^2)^{5/6}}
& \qquad
v_y &= {} \frac{x\cos\Big(\frac{1}{3}\arctan\big(\frac{y}{x}\big)\Big) +
y\sin\Big(\frac{1}{3}\arctan\big(\frac{y}{x}\big)\Big)}
{3(x^2 + y^2)^{5/6}}\\
u_y
&= \frac{y\cos\Big(\frac{1}{3}\arctan\big(\frac{y}{x}\big)\Big) -
x\sin\Big(\frac{1}{3}\arctan\big(\frac{y}{x}\big)\Big)}
{3(x^2 + y^2)^{5/6}}
& \qquad
v_y &= {} \frac{-y\cos\Big(\frac{1}{3}\arctan\big(\frac{y}{x}\big)\Big) +
x\sin\Big(\frac{1}{3}\arctan\big(\frac{y}{x}\big)\Big)}
{3(x^2 + y^2)^{5/6}}
\end{alignat*}
Therefore, (2) is satisfied. (1) is satisfied since \(f\) is infinitely differentiable so the first partial exist. How can I determine the domain D that f is analytic in?

Is it \(\mathbb{C}-\{0\}\)?

 

[mathbalarka]

The principal branch of the function is *not* everywhere analytic.

Hint $z^{1/3} = \exp(1/2 \cdot \log(z))$, and the principal branch of log is analytic (prove it) everywhere except $z = 0$ and the branch cut $[-\infty, 0]$. $\exp(z)$ is everywhere analytic (prove it). Composition of analytic functions are analytic.

 

[Dustinsfl]

The principal branch of the function is *not* everywhere analytic.

Hint $z^{1/3} = \exp(1/2 \cdot \log(z))$, and the principal branch of log is analytic (prove it) everywhere except $z = 0$ and the branch cut $[-\infty, 0]$. $\exp(z)$ is everywhere analytic (prove it). Composition of analytic functions are analytic.

I am aware that \(z^{1/3} = \exp(1/3 \cdot \ln(z)) = \exp(1/3\ln\lvert z\rvert + i(\arg(z) + 2\pi k)) = \lvert z\rvert^{1/3}e^{i/3(\arg(z) + 2\pi k)}\)

So since we have a branch cut on \((-\infty, 0]\), would we say the function is analytic on the domain \(\mathbb{C} - \{(-\infty, 0]\}\)?

 

[mathbalarka]

Yes, as $\log(z)$ is analytic on that region and $\exp(z)$ is entire.

 

[Dustinsfl]

Yes, as $\log(z)$ is analytic on that region and $\exp(z)$ is entire.

How can I show (3) and (4) apply on this domain D?

 

[Prove It]

I want to show \(f(z) = z^{1/3}\) is analytic. I need to show that for some domain D:
(1) the partial derivatives exist in D
(2) the C-R equations are satisfied
(3) f is continuous in D
(4) the partial derivatives are continuous in D
Then f is analytic in D.
\begin{alignat*}{3}
u_x
&= \frac{x\cos\Big(\frac{1}{3}\arctan\big(\frac{y}{x}\big)\Big) +
y\sin\Big(\frac{1}{3}\arctan\big(\frac{y}{x}\big)\Big)}
{3(x^2 + y^2)^{5/6}}
& \qquad
v_y &= {} \frac{x\cos\Big(\frac{1}{3}\arctan\big(\frac{y}{x}\big)\Big) +
y\sin\Big(\frac{1}{3}\arctan\big(\frac{y}{x}\big)\Big)}
{3(x^2 + y^2)^{5/6}}\\
u_y
&= \frac{y\cos\Big(\frac{1}{3}\arctan\big(\frac{y}{x}\big)\Big) -
x\sin\Big(\frac{1}{3}\arctan\big(\frac{y}{x}\big)\Big)}
{3(x^2 + y^2)^{5/6}}
& \qquad
v_y &= {} \frac{-y\cos\Big(\frac{1}{3}\arctan\big(\frac{y}{x}\big)\Big) +
x\sin\Big(\frac{1}{3}\arctan\big(\frac{y}{x}\big)\Big)}
{3(x^2 + y^2)^{5/6}}
\end{alignat*}
Therefore, (2) is satisfied. (1) is satisfied since \(f\) is infinitely differentiable so the first partial exist. How can I determine the domain D that f is analytic in?

Is it \(\mathbb{C}-\{0\}\)?

If a function is analytic, doesn't that mean you can write it as a series that is convergent over a neighbourhood?

$\displaystyle \begin{align*} z^{\frac{1}{3}} = \left[ 1 + \left( z - 1 \right) \right] ^{\frac{1}{3}} \end{align*}$

and you can apply the Binomial series expansion.

 

[mathbalarka]

How can I show (3) and (4) apply on this domain D?

Well, it is clear that $\log(z)$ (the principal branch) is not continuous at $(-\infty, 0]$ (arg(z) is not continuous there) and thus is not analytic. Transform the C-R equations into polar coordinates as $(x, y) \mapsto (r\sin(\theta), r\cos(\theta))$ and set $z = x + iy$. The calculations are quite straightforward.

 

[Dustinsfl]

If a function is analytic, doesn't that mean you can write it as a series that is convergent over a neighbourhood?

$\displaystyle \begin{align*} z^{\frac{1}{3}} = \left[ 1 + \left( z - 1 \right) \right] ^{\frac{1}{3}} \end{align*}$

and you can apply the Binomial series expansion.

I don't see how we can write that as a power series.

 

[Dustinsfl]

Well, it is clear that $\log(z)$ (the principal branch) is not continuous at $(-\infty, 0]$ (arg(z) is not continuous there) and thus is not analytic. Transform the C-R equations into polar coordinates as $(x, y) \mapsto (r\sin(\theta), r\cos(\theta))$ and set $z = x + iy$. The calculations are quite straightforward.

\[
u_x = v_y = \frac{\cos(2/3\cdot\theta)}{3r^{2/3}}\\
u_y = -v_x = \frac{\sin(2/3\cdot\theta)}{3r^{2/3}}
\]
So the partial derivatives aren't continuous at the origin.

 

[mathbalarka]

We have already excluded out the branch cut at $(-\infty, 0]$. $z^{1/3}$ is not analytic in any region encircling $0$, in particular.

 

[Dustinsfl]

We have already excluded out the branch cut at $(-\infty, 0]$. $z^{1/3}$ is not analytic in any region encircling $0$, in particular.

That is the information learned from your suggestion on polar coordinates though.

 

[Prove It]

I don't see how we can write that as a power series.

Can't you? Look up the Binomial Series.

$\displaystyle \begin{align*} \left( 1 + x \right) ^{\alpha} = \sum_{k = 0}^{\infty}{ {\alpha \choose{k}} x^k } \end{align*}$

where $\displaystyle \begin{align*} { \alpha \choose{k}} = \frac{ \alpha \left( \alpha - 1 \right) \left( \alpha - 2 \right) \dots \left( \alpha - k + 1 \right) }{k!} \end{align*}$

In this case, $\displaystyle \begin{align*} x = z - 1 \end{align*}$ and $\displaystyle \begin{align*} \alpha = \frac{1}{3} \end{align*}$.