[proofplan] We use the completed Dirichlet $L$-function attached to a primitive character. For a primitive nonprincipal character, this completed function is entire, while its gamma factor has simple poles exactly at the points $s=-a-2k$. Since the remaining power factor is holomorphic and nonvanishing there, the only way the completed product can remain holomorphic is for $L(s,\chi)$ to vanish at each such point. [/proofplan] [step:Introduce the completed Dirichlet $L$-function and its analytic properties] Let \begin{align*} H: \mathbb{C} &\to \mathbb{C}\setminus\{0\} \\ s &\mapsto \left(\frac{q}{\pi}\right)^{(s+a)/2} \end{align*} denote the exponential factor in the completed $L$-function. Since $q/\pi>0$, the function $H$ is entire and has no zeros. Define the completed Dirichlet $L$-function \begin{align*} \Lambda(\cdot,\chi): \mathbb{C} &\to \mathbb{C} \\ s &\mapsto H(s)\,\Gamma\!\left(\frac{s+a}{2}\right)L(s,\chi). \end{align*} By the [analytic continuation](/page/Analytic%20Continuation) and functional equation theorem for primitive Dirichlet $L$-functions, applied to the primitive nonprincipal character $\chi$, the function $\Lambda(\cdot,\chi)$ is entire, and $L(\cdot,\chi)$ is entire. We record this as a cited result not yet in the wiki: primitive nonprincipal Dirichlet $L$-functions have entire completed $L$-functions. [/step] [step:Locate the poles of the gamma factor] Fix an integer $k \ge 0$, and define \begin{align*} s_0 := -a-2k. \end{align*} Then \begin{align*} \frac{s_0+a}{2}=\frac{-a-2k+a}{2}=-k. \end{align*} The gamma function $\Gamma$ has a simple pole at every nonpositive integer. Therefore the function \begin{align*} G: \mathbb{C}\setminus\{s_0\} &\to \mathbb{C} \\ s &\mapsto \Gamma\!\left(\frac{s+a}{2}\right) \end{align*} has a simple pole at $s_0$. More explicitly, since $\Gamma(z)$ has a simple pole at $z=-k$, the composition with the affine map $s \mapsto (s+a)/2$ has a simple pole at $s=s_0$. Thus there is a punctured neighbourhood $U\setminus\{s_0\}$ of $s_0$ and a [holomorphic function](/page/Holomorphic%20Function) \begin{align*} \psi: U &\to \mathbb{C} \end{align*} with $\psi(s_0)\ne 0$ such that \begin{align*} \Gamma\!\left(\frac{s+a}{2}\right)=\frac{\psi(s)}{s-s_0} \end{align*} for every $s \in U\setminus\{s_0\}$. [/step] [step:Force cancellation by the holomorphic factor $L(s,\chi)$] Restrict $U$ if necessary so that $H(s)\ne 0$ for every $s \in U$; this is possible because $H$ is entire and nonvanishing on all of $\mathbb{C}$. On $U\setminus\{s_0\}$ we have \begin{align*} \Lambda(s,\chi) =H(s)\Gamma\!\left(\frac{s+a}{2}\right)L(s,\chi) =\frac{H(s)\psi(s)L(s,\chi)}{s-s_0}. \end{align*} The function \begin{align*} M: U &\to \mathbb{C} \\ s &\mapsto H(s)\psi(s)L(s,\chi) \end{align*} is holomorphic on $U$, because $H$, $\psi$, and $L(\cdot,\chi)$ are holomorphic there. Since $\Lambda(\cdot,\chi)$ is entire, it is holomorphic at $s_0$. The displayed formula shows that the possible principal part of $\Lambda(\cdot,\chi)$ at $s_0$ is determined by $M(s_0)/(s-s_0)$. Holomorphicity of $\Lambda(\cdot,\chi)$ at $s_0$ therefore forces \begin{align*} M(s_0)=0. \end{align*} But \begin{align*} M(s_0)=H(s_0)\psi(s_0)L(s_0,\chi). \end{align*} Here $H(s_0)\ne 0$ and $\psi(s_0)\ne 0$, so division by these two nonzero complex numbers gives \begin{align*} L(s_0,\chi)=0. \end{align*} Substituting $s_0=-a-2k$ yields \begin{align*} L(-a-2k,\chi)=0. \end{align*} Since $k\ge 0$ was arbitrary, the claimed vanishing holds for every integer $k\ge 0$. When $a=0$ and $k=0$, this gives $L(0,\chi)=0$. [/step]