Argument of a complex number confusion - LaGrange's method and root finding I am trying to use Lagrange's method to find the cubic roots of a polynomial with real coefficients. This involves taking the cubic root of a complex number and selecting the correct root (one of the several possible roots is appropriate, depending on certain conditions).
For instance, consider this case:
$arg\sqrt[3]x=\frac {2\pi}{3} + \frac 1 3 arg (x)$ if $arg(x)=\pi$
Ultimately, I need to calculate $\sqrt[3]x$ for some known value of $x$. For instance, $x=-1$. In that case, $arg(-1)=\pi$.
At the moment, I'm trying to replicate the results from this paper:
Ting Zhao, Dongming Wang, Hoon Hong.  Solution Formulas for Cubic Equations Without or With Constraints.  Journal of Symbolic Computation, Elsevier, 2011, 46 (8), pp.904-918.
The authors give the following example:
$\sqrt[3]{-1}=\sqrt[3]{e^{i\pi}}=e^{i\pi}=-1$
This example is very mysterious to me. From Wikipedia, I found the identity $x=\lvert x \rvert e^{i Arg(x)}$ but it seems that some steps are missing and I don't follow the logic. Clearly, the authors used this identity under the radical... what confuses me is that in the case I've given (and that they are using in this example) we had $arg\sqrt[3]{x}$. However, it seems like the argument has disappeared. Why is that? Is that correct?
I would like to see a step by step example of how to calculate $\sqrt[3]-1$ using the example case I've provided. Thanks!
 A: The notation for the real cube root and the complex cube root are, unfortunately, identical.  For the real cube root function, $\sqrt[3]{-1} = -1$.  For the complex cube root relation, $\sqrt[3]{-1} = \{-1, \frac{1}{2} - \frac{\sqrt{3}}{2}\mathrm{i}, \frac{1}{2} + \frac{\sqrt{3}}{2}\mathrm{i}\}$, taking $-1$ to its three cube roots.  In order to turn the complex cube root relation into a function, we need a convention for choosing which of the three cube roots is returned by the function.
The paper you cite specifies their convention on p. 906.  In that place, we see that they choose the argument of various complex quantities appearing in a cube root in the interval $(-\pi, \pi]$.  (This is the full range of "$\arg x$"s in the pieces of the piecewise function.  Any other $\arg x$ must be manipulated to be in this range.)  This means that $\sqrt[3]{-1} = \sqrt[3]{\mathrm{e}^{\mathrm{i}\pi}}$.  There are actually infinitely many choices for the argument, which I discuss below the horizontal line, below.
Since $\arg(-1) = \pi$ in their convention, we use the fifth choice in their piecewise definition of the argument of the cube root on that page:  $\arg \sqrt[3]{-1} = \frac{1}{3}\arg(-1) + \frac{2}{3}\pi = \frac{1}{3}\pi + \frac{2}{3}\pi$.  Of course, this, is just $\pi$.  We also know $|-1| = 1$ and $\sqrt[3]{1}$ has $\arg 0$ by their convention, so is using the real cube root.  Then $\sqrt[3]{1} = 1$ and we find
$$  |\sqrt[3]{-1}| = \sqrt[3]{|-1|} = \sqrt[3]{1}  = 1  \text{.}  $$
Therefore, the magnitude of $\sqrt[3]{-1}$ is $1$ and the argument is $\pi$.  I.e., by their argument convention, the cube root function gives
$$  \sqrt[3]{-1} = 1 \cdot \mathrm{e}^{\mathrm{i}\pi} = -1  \text{.}  $$

More generally, roots are a place where the standard, common notation hides a crucial fact.  Let's revisit the cube root that you recite,
$$  \sqrt[3]{-1} = \sqrt[3]{\mathrm{e}^{\mathrm{i}\pi}} = \mathrm{e}^{\mathrm{i}\pi} = -1  \text{,}  $$
but show the bit(s) that "everyone knows",
$$  \sqrt[3]{-1} = \sqrt[3]{\mathrm{e}^{\mathrm{i}\pi + 2\pi \mathrm{i}\Bbb{Z}}} = \mathrm{e}^{\mathrm{i}\pi/3 + 2\pi \mathrm{i}\Bbb{Z}/3} \ni -1  \text{.}  $$
So what's this "$2\pi\mathrm{i}\Bbb{Z}$" stuff about?  Notice that $\mathrm{e}^{2\pi\mathrm{i} k} = 1$ for any choice of $k$ in the integers, typically denoted "$\Bbb{Z}$".  So the notation in the second expression in the display above is reminding us that there are infinitely many different powers of $\mathrm{e}$ that have the value $-1$.  Then we take the cube root by dividing the exponent by $3$.
This gives us an infinite list of powers of $\mathrm{e}$, but maybe it only gives one complex number, just written many different ways.  We're not that lucky.  In fact, it gives all three complex cube roots of $-1$, which, really, is what we should expect the cube root operation to do.  Let's make a table for our different choices of the integer in $\Bbb{Z}$:
\begin{align*}
k&  &&|&  \mathrm{e}^{\mathrm{i}\pi/3 + 2\pi \mathrm{i}k/3}&  \\ \hline
 \vdots&  &&&  &\vdots  \\
 -2&  &&&  \mathrm{e}^{\mathrm{i}\pi/3 + 2\pi \mathrm{i}(-2)/3} &= \mathrm{e}^{-3\mathrm{i}\pi/3} = -1  \\
-1&  &&&  \mathrm{e}^{\mathrm{i}\pi/3 + 2\pi \mathrm{i}(-1)/3} &= \mathrm{e}^{-1\mathrm{i}\pi/3} = \frac{1}{2} - \frac{\sqrt{3}}{2}\mathrm{i}  \\
 0&  &&&  \mathrm{e}^{\mathrm{i}\pi/3 + 2\pi \mathrm{i}(0)/3} &= \mathrm{e}^{1\mathrm{i}\pi/3} = \frac{1}{2} + \frac{\sqrt{3}}{2}\mathrm{i}  \\
 1&  &&&  \mathrm{e}^{\mathrm{i}\pi/3 + 2\pi \mathrm{i}(1)/3} &= \mathrm{e}^{3\mathrm{i}\pi/3} = -1  \\
2&  &&&  \mathrm{e}^{\mathrm{i}\pi/3 + 2\pi \mathrm{i}(2)/3} &= \mathrm{e}^{5\mathrm{i}\pi/3} = \frac{1}{2} - \frac{\sqrt{3}}{2}\mathrm{i}  \\
\vdots&  &&&  &\vdots
\end{align*}
As we proceed through the integers, these values cycle through the three cube roots of $-1$.
The last bit of notation in the "show the bit(s)" display is the "$\ni$", which means $-1$ is one of the three values in the list of numbers produced by the cube root.
The paper you cite makes a choice of complex argument for each number you pass to a square or cube root.  Their choice has the nice property that it gives the roots of a cubic having real coefficients by following the usual equations, without having to try all the cube roots to see which ones actually do work and which ones don't.  So this convention on the argument is only useful for this one method of solving the cube root.
