Argument of a Gauss sum

It is well known that the Gauss sum of a Dirichlet character modulo $N$

$$G(\chi)=\sum_{a=1}^N\chi(a)e^{2\pi ia/N}$$

Moreover,$$\vert G(\chi)\vert=\sqrt{N}$$

when $\chi$ is primitive.

Question :If $G(\chi)=-\sqrt{N}$, the root number of L-function associated to the character $\chi$ is -1 or -i, which is not likely to occur. How to rule out such possibilities?

• The background to your question is incorrect: the functional equation relates $L(s,\chi)$ and $L(1-s,\overline{\chi})$, so you'd get an automatic zero at $s = 1/2$ from the root number being $-1$ only when $\chi = \overline{\chi}$, meaning $\chi$ is quadratic. For quadratic $\chi$ the root number is 1, so you learn nothing about a possible zero at $s=1/2$. And if $\chi$ is not quadratic then no value of the root number could force $L(1/2,\chi)$ to be $0$ since $\overline{\chi}$ is not the same character as $\chi$. Jun 29 '13 at 0:18
Here is a remark, if $\chi$ is a quadratic character. Then there is a remarkable result of Gauss, saying that $$G(\chi, N)=\begin{cases} \sqrt{N}, \;\text{ if } N\equiv 1 (4) \cr 0, \; \text{ if } N\equiv 2 (4) \cr i\sqrt{N} \; \text{ if } N\equiv 3 (4) \cr (1+i)\sqrt{N} \;\text{ if } N\equiv 0 (4) \end{cases}$$ So this rules out that $G(\chi, N)=-\sqrt{N}$. It took Gauss over $4$ years to prove this. A proof is in section $9.10$ of Tom Apostol's book on analytic number theory.
For characters of higher orders $k>4$, I believe it is an open problem, which roots of unity can occur. If $\chi$ is a real, primitive Dirichlet character, then only $1$ and $i$ can occur (i.e., $\sqrt{N}$ or $i\sqrt{N}$).
P.S.The corollary 3 of T.Funakura states that if $G(\chi)/\sqrt{N}$ is a fourth root of unity, then $\chi$ is a real character.