Given a lattice Γ ⊂ C , a Theta function ϑ : C → C is a holomorphic function with the following property: ϑ ( z + γ ) = e 2 i π a γ z + b γ ϑ ( z )for every γ ∈ Γ, and a γ , b γ ∈ C .Exercise: A Theta function never vanishes iff ϑ ( z ) = e p ( z ) with p ( z ) a polynomial of degree at most 2.Hint: The "only if" part is trivial. The hint is: show that log ⁡ ( ϑ ( z ) ) = O ( 1 + | z | 2 ). I tried to apply log on both sides, or derive one and two times, or everything I could have thought of. I don't get where the square comes from.

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Given a lattice   Γ  ⊂      C  , a Theta function   ϑ  :      C    →      C   is a holomorphic function with the following property:  ϑ  (  z  +  γ  )  =      e          2      i      π              a                  γ                    z      +              b                  γ                      ϑ  (  z  )for every   γ  ∈  Γ, and       a          γ        ,      b          γ        ∈      C  .Exercise: A Theta function never vanishes iff   ϑ  (  z  )  =      e          p      (      z      )       with   p  (  z  ) a polynomial of degree at most 2.Hint: The &quot;only if&quot; part is trivial. The hint is: show that   log  ⁡  (  ϑ  (  z  )  )  =  O  (  1  +      |    z            |        2    ). I tried to apply log on both sides, or derive one and two times, or everything I could have thought of. I don&#039;t get where the square comes from.

charringpq49u · Accepted Answer

Look at the derivative of   log  ⁡  ϑ, i.e. at the function  q  :  z  ↦                    ϑ        ′            (      z      )              ϑ      (      z      )        .By assumption,   q is an entire function, and from the property of   ϑ we find      ϑ    ′    (  z  +  γ  )  =  2  π  i      a    γ        e          2      π      i              a        γ            z      +              b        γ              ϑ  (  z  )  +      e          2      π      i              a        γ            z      +              b        γ                  ϑ    ′    (  z  )  ,whence  q  (  z  +  γ  )  =  2  π  i      a    γ    +  q  (  z  )for each   γ  ∈  Γ.From this relation deduce that   q is a polynomial of degree   ⩽  1. (Consider a fundamental parallelogram of   Γ, and a corresponding basis       ω    1    ,        ω    2   of   Γ.) Then, since we have       p    ′    =  q for   p such that   ϑ  (  z  )  ≡      e          p      (      z      )      , it follows that such a   p is a polynomial of degree   ⩽  2.

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