TITLE: How to derive this curious approximation to the cube root of $a + bi$?
QUESTION [7 upvotes]: In this Wikipedia article in Portuguese is given the following approximation for the cube root of a complex number $ c = a + bi$: 
$$ \sqrt[3]{c} \approx k\left ( \frac{29z^3 + 261z^2 + 255z + 22}{7z^3 + 165z^2 +324z+71} \right ) $$
where $ \sqrt[3]{c} = \sqrt[3]{k^3z},\quad $ $ \forall k, z\in \mathbb{C}, z \neq  0 $
This gives an approximation, say $p_1$, and then you can use this approximation as a new value of $k$ to get another better approximation $p_2$, and so on. 
The question is: how this approximation can be derived? I think that it's an application of Newton's method or some truncated series, but I don't know the details.
Also, what precise is the approximation? Does converges to the cube root for any initial values?

REPLY [4 votes]: Cube Root
We can start with the Taylor Series
$$
\sqrt[3]{1+x}=1+\frac x3-\frac{x^2}9+\frac{5x^3}{81}-\frac{10x^4}{243}+\frac{22x^5}{729}-\frac{154x^6}{6561}+O\!\left(x^7\right)\tag{1}
$$
then use the Extended Euclidean Algorithm to get the Padé Approximation
$$
\sqrt[3]{1+x}=\frac{81+135x+63x^2+7x^3}{81+108x+36x^2+2x^3}+O\!\left(x^7\right)\tag{2}
$$
Now we can simply substitute $x\mapsto x-1$ and get
$$
f(x)=\frac{2+30x+42x^2+7x^3}{7+42x+30x^2+2x^3}=\sqrt[3]{x}+O\!\left((x-1)^7\right)\tag{3}
$$
This approximation is nice in that it gives $\sqrt[\large3]{\frac1x}=\dfrac1{\sqrt[3]x}$. However, the maximum it can return is $\frac72$.
The sequence defined by $u_0=1$ and $u_{n+1}=u_nf\!\left(x/u_n^3\right)$, converges to $\sqrt[3]{x}$, and the fact that the error is $O\!\left((x-1)^7\right)$ means that if $u_n$ is accurate to $d$ places, $u_{n+1}$ should be accurate to $7d$ places. For example, for $x=8$, 
$$
\begin{align}
u_1&=1.981746273197444478247642227&\text{$2$ places}\\
u_2&=1.999999999999997579278630790&\text{$15$ places}\\
u_3&=2.000000000000000000000000000
\end{align}
$$

The Approximation in the Question
$$
\frac{29z^3+261z^2+255z+22}{7z^3+165z^2+324z+71}=\sqrt[3]{z}+O\left((z-1)^5\right)
$$
This will not converge as fast as the approximation given in $(3)$ (and uses larger coefficients). I am not sure how this approximation was derived.

Square Root
The same thing can be done to accelerate the square root iteration using
$$
f(x)=\frac{1+21x+35x^2+7x^3}{7+35x+21x^2+x^3}=\sqrt{x}+O\!\left((x-1)^7\right)
$$
The sequence $u_0=1$ and $u_{n+1}=u_nf\!\left(x/u_n^2\right)$ converges in the same manner to $\sqrt{x}$.