[proofplan] The statement is restricted to the nonreal-valued case, which is the genuinely complex-character case. We first verify that $\chi^2$ is still nonprincipal, and then study the auxiliary Euler product $G(s)=\zeta(s)^3L(s,\chi)^4L(s,\chi^2)$ for real $s>1$. Its logarithm has a prime expansion whose first-order coefficients are nonnegative, giving a uniform lower bound for $\log |G(s)|$ as $s \to 1^+$. If $L(1,\chi)=0$, then the zero of $L(s,\chi)^4$ dominates the pole of $\zeta(s)^3$, while $L(s,\chi^2)$ stays bounded, forcing $G(s)\to 0$ and hence $\log |G(s)|\to -\infty$, a contradiction. [/proofplan]

[step:Verify that the square of the nonreal character is nonprincipal] Since the theorem statement assumes that $\chi$ is not real-valued, no reduction to the real-character case is needed. Define the Dirichlet character \begin{align*} \psi: \mathbb{Z} &\to \mathbb{C} \\ n &\mapsto \chi(n)^2. \end{align*} We claim that $\psi$ is nonprincipal. If $\psi$ were principal modulo $q$, then for every integer $n$ with $\gcd(n,q)=1$ we would have $\chi(n)^2=1$, hence $\chi(n)\in\{-1,1\}$. For $\gcd(n,q)>1$ we have $\chi(n)=0$. Thus $\chi$ would be real-valued on all of $\mathbb{Z}$, contradicting the standing assumption. Hence $\chi^2=\psi$ is nonprincipal. [/step]

[step:Record boundedness of the nonprincipal $L$-functions near $s=1$] Let $\rho: \mathbb{Z}\to \mathbb{C}$ be any nonprincipal Dirichlet character modulo $q$. Define the partial-sum map \begin{align*} A_\rho: \mathbb{N} &\to \mathbb{C} \\ N &\mapsto \sum_{n=1}^{N} \rho(n). \end{align*} For $\operatorname{Re}(s)>1$, define the Dirichlet $L$-function associated to $\rho$ by \begin{align*} L_\rho: \{s\in\mathbb{C}:\operatorname{Re}(s)>1\} &\to \mathbb{C} \\ s &\mapsto \sum_{n=1}^{\infty}\frac{\rho(n)}{n^s}, \end{align*} and write $L(s,\rho):=L_\rho(s)$. Since $\rho$ is periodic modulo $q$ and nonprincipal, the sum over one complete residue system is zero: \begin{align*} \sum_{r=1}^{q} \rho(r)=0. \end{align*} Therefore $|A_\rho(N)|\le q$ for every $N\in\mathbb{N}$. For $N\in\mathbb{N}$ and complex $s$ with $\operatorname{Re}(s)>1$, summation by parts gives \begin{align*} \sum_{n=1}^{N} \frac{\rho(n)}{n^s} = A_\rho(N)N^{-s} + \sum_{n=1}^{N-1} A_\rho(n)\left(n^{-s}-(n+1)^{-s}\right). \end{align*} On every compact set $K\subset\{s\in\mathbb{C}:\operatorname{Re}(s)>0\}$, the terms on the right converge uniformly as $N\to\infty$, because $|A_\rho(n)|\le q$ and \begin{align*} \left|n^{-s}-(n+1)^{-s}\right| \le C_K n^{-1-\sigma_K}, \qquad \sigma_K:=\inf_{s\in K}\operatorname{Re}(s)>0, \end{align*} where $C_K>0$ is a constant depending only on $K$. Thus $L(s,\rho)$ extends holomorphically to $\operatorname{Re}(s)>0$, and in particular it is bounded on some closed disc centered at $1$. Applying this to $\rho=\chi$ and $\rho=\chi^2$, both $L(s,\chi)$ and $L(s,\chi^2)$ are holomorphic and bounded in a neighbourhood of $s=1$. [/step]

[step:Expand the logarithm of the auxiliary Euler product] Let $\zeta: \{s\in\mathbb{C}:\operatorname{Re}(s)>1\}\to\mathbb{C}$ denote the Riemann zeta function defined by \begin{align*} \zeta(s):=\sum_{n=1}^{\infty}\frac{1}{n^s}. \end{align*} For real $s>1$, define the auxiliary function \begin{align*} G: (1,\infty) &\to \mathbb{C} \\ s &\mapsto \zeta(s)^3 L(s,\chi)^4 L(s,\chi^2). \end{align*} The absolutely convergent Euler products for $\zeta(s)$, $L(s,\chi)$, and $L(s,\chi^2)$ give \begin{align*} G(s) = \prod_{p} (1-p^{-s})^{-3} (1-\chi(p)p^{-s})^{-4} (1-\chi(p)^2p^{-s})^{-1}, \end{align*} where the product is over all primes $p$. For real $s>1$, each Euler factor above is nonzero because $|p^{-s}|<1$, $|\chi(p)p^{-s}|<1$, and $|\chi(p)^2p^{-s}|<1$. The Euler products are absolutely convergent, so their logarithms may be expanded termwise using the logarithmic series \begin{align*} -\log(1-z)=\sum_{m=1}^{\infty}\frac{z^m}{m}, \qquad |z|<1. \end{align*} Applying this with $z=p^{-s}$, $z=\chi(p)p^{-s}$, and $z=\chi(p)^2p^{-s}$, absolute convergence permits summing over $p$ and $m$, and taking real parts yields \begin{align*} \log |G(s)| = \sum_{p}\sum_{m=1}^{\infty} \frac{3+4\operatorname{Re}(\chi(p)^m)+\operatorname{Re}(\chi(p)^{2m})}{m p^{ms}}. \end{align*} Separate the $m=1$ terms from the terms with $m\ge 2$. Since $|\chi(p)|\le 1$, the latter contribution is bounded in absolute value, uniformly for $1<s\le 2$, by \begin{align*} 8\sum_{p}\sum_{m=2}^{\infty}\frac{1}{p^m} \le 8\sum_{n=2}^{\infty}\sum_{m=2}^{\infty}\frac{1}{n^m} <\infty. \end{align*} Hence \begin{align*} \log |G(s)| = \sum_p \frac{3+4\operatorname{Re}\chi(p)+\operatorname{Re}(\chi(p)^2)}{p^s} + O(1) \end{align*} as $s\to 1^+$. [/step]

[step:Use positivity of the first prime coefficients to bound $\log |G(s)|$ below] For a prime $p$ with $p\nmid q$, the value $\chi(p)$ has modulus $1$, so there exists $\theta_p\in\mathbb{R}$ such that \begin{align*} \chi(p)=e^{i\theta_p}. \end{align*} Then \begin{align*} 3+4\operatorname{Re}\chi(p)+\operatorname{Re}(\chi(p)^2) &= 3+4\cos\theta_p+\cos(2\theta_p)\\ &= 3+4\cos\theta_p+2\cos^2\theta_p-1\\ &= 2(1+\cos\theta_p)^2 \ge 0. \end{align*} If $p\mid q$, then $\chi(p)=0$, so the same coefficient equals $3\ge 0$. Therefore every first-order prime coefficient is nonnegative. The expansion from the previous step gives \begin{align*} \log |G(s)| \ge -C \end{align*} for all real $s$ sufficiently close to $1$ with $s>1$, where $C>0$ is a constant coming from the uniformly bounded $m\ge 2$ part of the logarithmic expansion. [/step]

[step:Derive the contradiction from a hypothetical zero of $L(s,\chi)$ at $s=1$] Assume for contradiction that \begin{align*} L(1,\chi)=0. \end{align*} Since $L(s,\chi)$ is holomorphic near $1$, there exists a function $h$ holomorphic and bounded near $1$ such that \begin{align*} L(s,\chi)=(s-1)h(s). \end{align*} Also $L(s,\chi^2)$ is bounded near $1$ because $\chi^2$ is nonprincipal. Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. For real $1<s<2$, the elementary integral comparison gives \begin{align*} \zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s} \le 1+\int_1^\infty t^{-s}\,d\mathcal{L}^1(t) = 1+\frac{1}{s-1}. \end{align*} Since $0<s-1<1$, this implies $\zeta(s)\le 2(s-1)^{-1}$. Hence, with $C_1:=8$, \begin{align*} |\zeta(s)|^3 \le C_1(s-1)^{-3} \end{align*} for all real $1<s<2$. Since $h$ and $L(s,\chi^2)$ are bounded near $1$, there is a constant $C_2>0$ such that \begin{align*} |L(s,\chi)^4 L(s,\chi^2)| \le C_2(s-1)^4 \end{align*} for all real $s>1$ sufficiently close to $1$. Combining these estimates, \begin{align*} |G(s)| = |\zeta(s)|^3 |L(s,\chi)|^4 |L(s,\chi^2)| \le C_1C_2(s-1). \end{align*} Hence $G(s)\to 0$ as $s\to 1^+$, so \begin{align*} \log |G(s)|\to -\infty. \end{align*} This contradicts the lower bound $\log |G(s)|\ge -C$ obtained from the Euler-product logarithm. Therefore the assumption $L(1,\chi)=0$ is false, and \begin{align*} L(1,\chi)\ne 0. \end{align*} [/step]