The convert(expr, radical) calling sequence replaces RootOfs of polynomials by appropriate expressions in radical notation if possible.

The conversion can fail if Maple cannot find radical expressions for the roots or if the correct radical expression cannot be selected. If the conversion fails, the RootOf remains unchanged.

If a root of unity cannot be expressed in terms of radicals, it is converted to an equivalent expression involving sin and cos terms.

If the argument n is included, the n-th root with respect to the ordering used by solve is returned. Otherwise, the following rules are applied to choose the root.

Binomials

An indexed RootOf of the form RootOf(P(_Z), index=i), where $P\left(\mathrm{\_Z}\right)=A{\mathrm{\_Z}}^m+B$ for some integer m is replaced by (-B/A)^(1/m)*(-1)^(2*(i-1)/m). If no index is given, then $i=1$ is assumed and ${\left(-\frac{B}{A}\right)}^{\frac{1}{m}}$ is returned. In particular, RootOf(_Z^2+1) and RootOf(_Z^2+1, index=1) are transformed into $\mathrm{I}$.

If the argument n is included, indexed RootOf can be converted to radical only if n equals the index $i$. Otherwise, an error is generated.

Other cases

Labeled $\mathrm{RootOf}$s are converted by the same rules as unlabeled.

If a $\mathrm{RootOf}$ can be evaluated numerically by using evalf, then the radical expression with the closest numerical approximation is returned, which is the value of $\mathrm{RootOf}$ with index=1. If numerical evaluation fails, for example, if the $\mathrm{RootOf}$ has symbolic coefficients, then the cause is one of the following.

The $\mathrm{RootOf}$ has only one argument. That is, it has the form RootOf(P(_Z)). The first radical expression for the ordering used by the solve command is returned.

The $\mathrm{RootOf}$ is indexed. The conversion usually fails because Maple is unable to find a radical expression equal to the input $\mathrm{RootOf}$ for all values of the parameters.

To a limited extent, the RootOf notation can be restored by using convert/RootOf.

If the argument of a trigonometric function is of the form $\frac{n\mathrm{\pi }}{120}$ where n is an integer, then Maple converts the function to radical form.

$\mathrm{convert}\left(\mathrm{RootOf}\left({\mathrm{\_Z}}^3-2\,\mathrm{index}=1\right)\,\mathrm{radical}\right)$

$2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

$\mathrm{convert}\left(\mathrm{RootOf}\left({\mathrm{\_Z}}^3-2\,\mathrm{index}=2\right)\,\mathrm{radical}\right)$

$2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(\mathrm{-1}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

$\mathrm{convert}\left(\mathrm{RootOf}\left({\mathrm{\_Z}}^3-2\,\mathrm{index}=2\right)\,\mathrm{radical}\,2\right)$

$-\frac{2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{2}+\frac{\mathrm{I}\sqrt{3}{2}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{2}$

$\mathrm{convert}\left(\mathrm{RootOf}\left({\mathrm{\_Z}}^3-2\,\mathrm{index}=2\right)\,\mathrm{radical}\,3\right)$

Error, (in `convert/radical`) indexed RootOf can only be converted to the same root number

$\mathrm{convert}\left(\mathrm{RootOf}\left({\mathrm{\_Z}}^3-2\,1.2\right)\,\mathrm{radical}\right)$

$2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

$\mathrm{convert}\left(\mathrm{RootOf}\left({\mathrm{\_Z}}^2+1\,\mathrm{index}=2\right)\,\mathrm{radical}\right)$

$\mathrm{-I}$

$\mathrm{convert}\left(\mathrm{RootOf}\left({\mathrm{\_Z}}^2+1\right)\,\mathrm{radical}\right)$

$\mathrm{I}$

$\mathrm{convert}\left(\mathrm{RootOf}\left({\mathrm{\_Z}}^3+\mathrm{\_Z}+1\,\mathrm{index}=1\right)\,\mathrm{radical}\right)$

$\frac{{\left(108+12\sqrt{93}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{12}-\frac{1}{{\left(108+12\sqrt{93}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}-\frac{\mathrm{I}\sqrt{3}\left(-\frac{{\left(108+12\sqrt{93}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{6}-\frac{2}{{\left(108+12\sqrt{93}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)}{2}$

$\mathrm{convert}\left(\mathrm{RootOf}\left({\mathrm{\_Z}}^3+\mathrm{\_Z}+1\right)\,\mathrm{radical}\right)$

$\frac{{\left(108+12\sqrt{93}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{12}-\frac{1}{{\left(108+12\sqrt{93}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}-\frac{\mathrm{I}\sqrt{3}\left(-\frac{{\left(108+12\sqrt{93}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{6}-\frac{2}{{\left(108+12\sqrt{93}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)}{2}$

$\mathrm{convert}\left(\mathrm{RootOf}\left({\mathrm{\_Z}}^3+\mathrm{\_Z}+x\right)\,\mathrm{radical}\right)$

$\frac{{\left(-108x+12\sqrt{81x^2+12}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{6}-\frac{2}{{\left(-108x+12\sqrt{81x^2+12}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}$

$\mathrm{convert}\left(\mathrm{RootOf}\left({\mathrm{\_Z}}^3+\mathrm{\_Z}+x\,\mathrm{index}=1\right)\,\mathrm{radical}\right)$

$\mathrm{RootOf}\left({\mathrm{\_Z}}^3+\mathrm{\_Z}+x\,\mathrm{index}=1\right)$

$\mathrm{convert}\left(\mathrm{RootOf}\left({\mathrm{\_Z}}^5+\mathrm{\_Z}+3\,\mathrm{index}=1\right)\,\mathrm{radical}\right)$

$\mathrm{RootOf}\left({\mathrm{\_Z}}^5+\mathrm{\_Z}+3\,\mathrm{index}=1\right)$

$\mathrm{convert}\left(\mathrm{RootOf}\left({\mathrm{\_Z}}^6+{\mathrm{\_Z}}^5+{\mathrm{\_Z}}^4+{\mathrm{\_Z}}^3+{\mathrm{\_Z}}^2+\mathrm{\_Z}+1\,\mathrm{index}=1\right)\,\mathrm{radical}\right)$

$\cos \left(\frac{2\mathrm{\pi }}{7}\right)+\mathrm{I}\sin \left(\frac{2\mathrm{\pi }}{7}\right)$

$\sin \left(\frac{\mathrm{\pi }}{30}\right)$

$\sin \left(\frac{\mathrm{\pi }}{30}\right)$

$\mathrm{convert}\left(\,'\mathrm{radical}'\right)$

$\frac{\sqrt{2}\sqrt{3}\sqrt{5-\sqrt{5}}}{8}-\frac{1}{8}-\frac{\sqrt{5}}{8}$

$\mathrm{convert}\left(\sec \left(\frac{5}{24}\mathrm{\pi }\right)\,'\mathrm{radical}'\right)$

$\frac{4}{\sqrt{8+2\sqrt{6}-2\sqrt{2}}}$

$\mathrm{convert}\left(\cot \left(\frac{323}{20}\mathrm{\pi }\right)\,'\mathrm{radical}'\right)$

$\frac{\left(\sqrt{5}-1\right)\left(4+\sqrt{2}\sqrt{5-\sqrt{5}}\right)}{4}$