An apparently new method to compute the $n$th root of any complex number

Question

I found  a series of articles (in Portuguese) by a Brazilian mathematician named Ludenir Santos, where presents a series of iterative methods, he said new, to extract nth roots of any complex number different from zero (and hence for real numbers), with any desired degree of accuracy. 

In his first articles, he has two different methods to arithmetically extract square and cube roots of complex numbers. Recently, he published an article that presents a general method for calculating nth roots of complex numbers, where $ n \in \mathbb{N}$. Here's the method:

Let $c = a+ib$. Then

$(1) \sqrt[n]{c} \approx k(1+\frac{x}{n}-\frac{x^2}{2n})$

where $x = \frac{c}{k^n}-1$ and $k$ is an approximation to the nth root of $c$. In the Wikipedia article in Portuguese, $k$ is taken as follow:

$|a| + |b| = m$. Then, $k = p$, where $ p^n$ is the nearest $n$th power to $m$. The sign of $k$ is defined as follow: 

If $|a| > |b|$, then $sign(k) = sign(a)$, and vice versa.

This formula really works and, according to the author, provides about 6 places correct after 3 iterations. In my view, however, the notation is not very successful, since, at first glance, seems as an approximation, when in fact it is an iterative  method: in the first iteration, we get a value, namely $k_0$, and this approximation is used as a new guess. This gives a better approximation $k_1$, and so on. The author clarify this point, but it is not immediately obvious because the notation used.

 The way he presents these results, however, more reminiscent of a  17th century mathematical paper than one of the 21th century, since it offers no demonstration of the method, or an indication of how it was obtained. This omission seems deliberate, since, as the author says here: "In 2014 , I discovered a really new area in mathematics, which allows me to make the formulas that come divulging ". I do not know what degree of truth that has this statement,since he published nothing about the alleged "new area" of mathematics.

However, the method really works, and it seems interesting for its simplicity and generality. So, my inquire are: how can prove the formula $(1)$?

Is a new method, as the author says ? I've never seen elsewhere, but don't have a lot of knowledge on this topic. I even read books that seem to deny the existence of such a method. For example, in Text Book On Algebra, p. 108, we read: "Now, there is no general arithmetical process for extracting the cube root of complex numbers and condequently Cardan's method fail to give the solution of a cubic all of whose roots are real and different". Depending on what is meant by " arithmetical process ", this formula shows that such a method does exist, or one like that Newton's method

So, my other question is: for calculating nth roots of complex numbers, to the present, what methods could be used for this purpose, apart from using trigonometry -that's to say, De Moivre's theorem- ?

Answer 1

If one allows successive approximation methods (which indeed should be fair game for numerical solutions), then Newton-Raphson has been known for a long time already.

The iteration you quote seems to be somewhat more work than using Newton-Raphson to find a root of $x^n−c$. In fact if you omit the $\frac{x^2}{2n}$ term, then it reduces to exactly the same iteration as Newton-Raphson.

The exact provenance of the $\frac{x^2}{2n}$ term we may never know for sure; at least it doesn't seem to harm the convergence over plain Newton-Raphson.

The use of the auxiliary variable $x=\frac{c}{k^n}-1$ suggests that the intention of the method would be to find a root (for $k$) of $\frac{c}{k^n}-1$. Using the Newton-Raphson formula on that would give $$ k \leftarrow k\left(1+\frac1n\left(\frac{x}{x+1}\right)\right) $$ where your paper has $$ k \leftarrow k\left(1+\frac1n\left(x-\tfrac12x^2\right)\right) $$

Now, $\frac{x}{x+1}$ and $x-\tfrac12x^2$ agree to first degree, which should be close enough to preserve good convergence, but the series expansion of $\frac{x}{x+1}$ is $x-x^2+O(x^3)$ -- so this is probably not exactly how the $\frac{x^2}{2n}$ term was invented. But something like that has likely been going on.

The way he presents these results, however, more reminiscent of a 17th century mathematical paper than one of the 21th century, since it offers no demonstration of the method, or an indication of how it was obtained.

That seems suitable, since what he has rediscovered might well be pretty close to the nascent calculus of the late 17th century.

Answer 2

If $|x|<1$ then the binomial series for the n-th root reads as $$ (1+x)^{\frac1n}=1+\frac1n x+\frac{\frac1n(\frac1n-1)}2x^2+… $$ which shows that the given formula is an approximation to the truncation after the quadratic term, replacing $\frac{n-1}n$ with $1$. Using the non-approximated truncation should result in 3rd order convergence.

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One can transform the quadratic Taylor polynomial into a fraction of linear terms while retaining the approximation order, \begin{align} 1+\frac xn\left(1-\frac{(n-1)x}{2n}\right)+O(x^3) &=1+\frac xn\left(1+\frac{(n-1)x}{2n}\right)^{-1}+O(x^3)\\ &=\frac{2n+(n-1)x+2x}{2n+(n-1)x}+O(x^3)\\ &=\frac{2n+(n+1)x}{2n+(n-1)x}+O(x^3) \end{align} Rewritten with $x=z-1$ this gives the fraction $$ F_n(z)=\frac{(n-1)+(n+1)z}{(n+1)+(n-1)z} $$ which in the iteration formula $$ k_+=k·F_n(\frac c{k^n}) $$ recovers Halley's method applied to the $n$th root. Recall that Halley's method for $f(k)=k^n-c$ is Newtons method for $$g(k)=\frac{f(k)}{\sqrt{|f'(k)|/n}}=k^{(n+1)/2}-c·k^{-(n-1)/2}.$$

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For an order 5 method the Pade approximant of $\sqrt[n]z+O((z-1)^5)$ would be $$\scriptsize F_n(z)=\frac{(n+1)(2n+1)z^2+ 2(2n+1)(2n-1)z+(n-1)(2n-1)}{(n-1)(2n-1)z^2+ 2(2n+1)(2n-1)z+(n+1)(2n+1),} $$ For an order 7 method the Pade approximant is $$\scriptsize F_n(z)=\frac{(n+1)(2n+1)(3n+1)z^3+3(2n+1)(3n+1)(3n-1)z^2+3(2n-1)(3n+1)(3n-1)z+(n-1)(2n-1)(3n-1)}{(n-1)(2n-1)(3n-1)z^3+3(2n-1)(3n+1)(3n-1)z^2+3(2n+1)(3n+1)(3n-1)z+(n+1)(2n+1)(3n+1)} $$