Branch of multiple-valued function (complex analysis) Find a branch of $(z^3 - 1)^{1/3}$ which is analytic in $|z| &gt; 1$


So we essentially want to study $\frac 13\text{Log} (1 - \frac 1{z^3} )$, the principal branch of the logarithm where $-\pi &lt; \text{Arg} (z) &lt; \pi $. Problems arise when $\text{Re}(z^3 - 1) \leq 0 $ and $\text{Im} (z^3 - 1) = 0$ i.e. when $z^3 - 1$ is a non-positive, real number.

Setting these equations up, how can I produce the result $|z| &gt; 1$?
Let
$$
\omega_1=1, \quad \omega_2=\frac{-1+i\sqrt{3}}{2}\quad\text{and}\quad
\omega_3=\frac{-1-i\sqrt{3}}{2}.
$$

A domain for $\,\sqrt[3]{1-z^3}\,$ could be 
$$
\Omega=\mathbb C\smallsetminus \big([0,\omega_1]\cup[0,\omega_2]\cup [0,\omega_3]\big).
$$
By $[a,b]$, for $a,b\in \mathbb C$ we denote the segment connecting $a$ and $b$ in the complex plane.

Clearly $\{z:\lvert z\rvert&gt;1\}\subset\Omega$.

In order to prove it you need the following lemma:

If $a,b$ belong to the same connected component of $\mathbb C\smallsetminus\Omega$, then
$$
f(z)=\log\left(\frac{z-a}{z-b}\right),
$$
is definable as holomorphic in $\Omega$.

Using this lemma, we can easily see that $\omega_1,\omega_2,\omega_3$ belong to the one and only component of $\mathbb C\smallsetminus\Omega$ and
$$
\sqrt[3]{1-z^3}=(z-\omega_3)\exp\left(
\frac{1}{3}\log\left(\frac{z-\omega_1}{z-\omega_3}\right)+
\frac{1}{3}\log\left(\frac{z-\omega_2}{z-\omega_3}\right)
\right).
$$
Let
$$
\omega_1=1, \quad \omega_2=\frac{-1+i\sqrt{3}}{2}\quad\text{and}\quad
\omega_3=\frac{-1-i\sqrt{3}}{2}.
$$

A domain for $\,\sqrt[3]{1-z^3}\,$ could be 
$$
\Omega=\mathbb C\smallsetminus \big([0,\omega_1]\cup[0,\omega_2]\cup [0,\omega_3]\big).
$$
By $[a,b]$, for $a,b\in \mathbb C$ we denote the segment connecting $a$ and $b$ in the complex plane.

Clearly $\{z:\lvert z\rvert&gt;1\}\subset\Omega$.

In order to prove it you need the following lemma:

If $a,b$ belong to the same connected component of $\mathbb C\smallsetminus\Omega$, then
$$
f(z)=\log\left(\frac{z-a}{z-b}\right),
$$
is definable as holomorphic in $\Omega$.

Using this lemma, we can easily see that $\omega_1,\omega_2,\omega_3$ belong to the one and only component of $\mathbb C\smallsetminus\Omega$ and
$$
\sqrt[3]{1-z^3}=(z-\omega_3)\exp\left(
\frac{1}{3}\log\left(\frac{z-\omega_1}{z-\omega_3}\right)+
\frac{1}{3}\log\left(\frac{z-\omega_2}{z-\omega_3}\right)
\right).
$$