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Let $f(z) = \sum_{n=1}^\infty A(n)n^{\frac{k-1}{2}}e(nz)$ be a modular form of weight $k$ for a half integer $k$. Put $$L(s,f) = \sum_{n=1}^\infty \frac{A(n)}{n^s} $$ to be the $L$-function.

Further assume that $f$ is an eigenfunction of the half integral weight Hecke operators.

Has there been located any zeros of this $L$ function, for any choice of $f$, in the critical strip which are not on the critical line $\Re(s) = \frac 12$.

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Have you done any computations yourself? While I'm dubious that this should be true for almost any form, it's worth noting that $L(s,\theta_\chi)=L(2s-1/2,\chi)$ for a non-trivial Dirichlet character $\chi$, so RH presumably holds in this case. In general, though, the multiplicative structure of half-integral weight eigenforms is more complex, and I'd be very surprised if it were to hold if the form is orthogonal to the space of unary theta functions. –  rlo Apr 17 '13 at 3:08
    
No I haven't done any computations, but how would one go about computing the zeros of such a modular form. I agree that for theta functions something special must be happening. –  Eren Mehmet Kiral Apr 19 '13 at 21:40

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