# How to solve fifth-degree equations by elliptic functions?

From F. Klein’s books, It seems that one can find the roots of a quintic equation

$$z5+az4+bz3+cz2+dz+e=0z^5+az^4+bz^3+cz^2+dz+e=0$$

(where $$a,b,c,d,e∈Ca,b,c,d,e\in\Bbb C$$) by elliptic functions. How to get that?

To solve the general quintic using elliptic functions, one way is to reduce it to Bring-Jerrard form,

a transformation which can be done in radicals. (See this post.) To solve $(1)$, define,

with the complete elliptic integral of the first kind $K(k)$ and elliptic parameter $m=k^2$ (with $p$ also given in Mathematica syntax above).

Method 1: For $j=0,1,2,3,4$, let,

where $\eta(\tau)$ is the Dedekind eta function, then,

with the sign chosen appropriately.

$\color{blue}{\text{Remark}}$: The five roots $x_n$can be found by using $p_n = i\frac{K(k')}{K(k)}+16n$ for $n = 0,1,2,3,4$.

Method 2: For $j=0,1,2,3,4$, let,

with $\vartheta_n(0,q)$ as the Jacobi theta functions, then,

and the sign of $\zeta_8$ chosen appropriately. Note that $(S_j)^8 = (T_j)^8$.

$\color{blue}{\text{Remark}}$: One can also find the other roots $x_i$, but is not as simple as in Method 1. (As one can see, you need much more than radicals to solve the general quintic.)

Example. Let,

so $d = \pm1$. Plugging it into $(2)$ gives $k \approx 0.072696$, and $m=k^2$, so $p \approx 2.550572\,i$. Since both methods use a square root, using the formulas one eventually finds,

$\color{blue}{\text{Afterword}}$ (Added June 2015): Given the nome $q = \exp(i \pi \tau)$, the two methods involve the function known either as the modular lambda function or elliptic lambda function, $\lambda(\tau)$ which has a beautiful q-continued fraction,

studied by Ramanujan (who also had his own method to solve solvable quintics).