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174    9. Elliptic and Hypergeometric Functions

        Exercises

         1. Show that ζ(z + ω) = ζ(z) + η(ω), where η(ω) are constants.
         2. For η(ω i ) = η i derive Legendre’s relation

                                    η 1 ω 2 − η 2 ω 1 = 2πi

            by considering  ζ(z)dz, where P is a fundamental parallelogram whose boundary is
                        ∂ P
            disjoint from L.
         3. Show that the following infinite product converges and represents an entire function:
                                                        z
                                          z     )  z  1      2  *
                         σ(z) = z      1 −   · exp  +        .
                                          ω       ω   2 ω
                                ω∈L−0
            Show also that ζ(z) = σ (z)/σ(z) = (d/dz) log σ(z).

         4. For L = Zω 1 + Zω 2 and ζ(z + ω i ) = ζ(z) + η i show that
                                              "         #
                                                     ω i
                             σ (z + ω i ) =−σ(z)exp η i z +  .
                                                      2
         5. Show that any elliptic function can be written as
                                          σ(z − a i )
                                       n
                                     c           .
                                          σ(z − b i )
                                      i=1
            where c is a constant.

        §4. The Differential Equation for ℘(z)

                                                    2


        Since ℘ (z) is an odd elliptic function, its square ℘ (z) is an even elliptic function,
        and by (3.3) it is a rational function in ℘(z). In this section we will prove that this
        rational function is a cubic polynomial. To do this, we consider the Laurent develop-
        ment of ζ(z, L), ℘(z, L),and ℘ (z, L) at the origin.

           From the geometric series we have
                                      1       1        n
                                                     z
                                          =−
                                    z − ω     ω      ω
                                                0≤n
           and
                             1     1    z     1        n
                                                     z
                                +    +    =−
                           z − ω   ω   ω 2    ω      ω
                                                2≤n
        which converge for |z|≤|ω|. Thus
                                          )               *
                                 1           1     1    z
                        ζ(z, L) =  +             +   +
                                 z         z − ω   ω   ω 2
                                     ω∈L−0
                                 1             z n
                               =   −
                                 z           ω n+1
                                     ω∈L−0,2≤n
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