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

z5+az4+bz3+cz2+dz+e=0

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

Update: How to transform a general higher degree five or higher equation to normal form?

**Answer**

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

x5−x+d=0

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

k=tan(14arcsin(1625√5d2))

p=iK(k′)K(k)=iEllipticK[1-m]EllipticK[m]

with the *complete elliptic integral of the first kind* K(k) and *elliptic parameter* m=k2 (with p also given in *Mathematica* syntax above).

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

Sj=uj√2η(τj)η2(4τj)η3(2τj)

τj=110(p+2j)

u=ζ8=exp(2πi/8)

S5=√2η(5p2)η2(10p)η3(5p)

where η(τ) is the *Dedekind eta function*, then,

x=±12⋅53/4(k2)1/8√k(1−k2)(S0+S5)(S1+iS4)(iS2+S3)

with the sign chosen appropriately.

Remark: The five roots xncan be found by using pn=iK(k′)K(k)+16n for n=0,1,2,3,4.

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

Tj=(ϑ2(0,wjq1/5)ϑ3(0,wjq1/5))1/2

q=exp(iπp0)

w=ζ5=exp(2πi/5)

T5=q5/8(q5)1/8(ϑ2(0,q5)ϑ3(0,q5))1/2

with ϑn(0,q) as the *Jacobi theta functions*, then,

x=±ζ82⋅53/4(k2)1/8√k(1−k2)(T0+T5)(T1−iT4)(T2+T3)

and the sign of ζ8 chosen appropriately. Note that (Sj)8=(Tj)8.

Remark: One can also find the other roots xi, 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,

x5−x+1=0

so d=±1. Plugging it into (2) gives k≈0.072696, and m=k2, so p≈2.550572i. Since both methods use a square root, using the formulas one eventually finds,

x=∓1.1673039…

Afterword (Added June 2015): Given the *nome* q=exp(iπτ), the two methods involve the function known either as the *modular lambda function* or *elliptic lambda function*, λ(τ) which has a beautiful q-continued fraction,

(λ(τ))1/8=√2η(τ2)η2(2τ)η3(τ)=(ϑ2(0,q)ϑ3(0,q))1/2=√2q1/81+q1+q+q21+q2+q31+q3+⋱

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

**Attribution***Source : Link , Question Author : ziang chen , Answer Author : Community*