Range of convergence for a Gaussian in the complex plane

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Given the following integral

$$Z(\lambda) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} dx \exp\left( -\frac{x^2}{2!}-\frac{\lambda}{4!}x^4\right)$$

We want to find for what values of $\lambda$ does this integral converge. If we assume that $\lambda \in \Bbb R$ we see three clear cases to inspect

  • $\lambda < 0$

Given that the $x^4$ dominates as $x\to \infty$, we conclude that $Z(\lambda)$ diverges.

  • $\lambda = 0$

We simply get the Gaussian integral, so $Z(\lambda)$ converges.

  • $\lambda > 0$

Given that the $x^4$ dominates as $x\to \infty$, we see that such term decreases faster that the $x^2$, which we know that converges. So by comparison, we conclude that $Z(\lambda)$ converges.

Hence we determine that $Z(\lambda)$ converges for $\lambda \geq 0$

But what if we assume $\lambda \in \Bbb C$. What cases should we consider? I guess we would follow an analogous procedure but I do not see it...