Equality of Integrals When Shifting Bounds By Complex Constant

Question

I am trying to determine whether the following is true: Let $f(x)$ be any integrable function satisfying $$\lim_{x\to\infty}f(x) =0$$ Then, for any $k\in \mathbb{C}$, we have $$\int_{0}^{\infty} f(x) dx=\int_{0}^{\infty + k} f(x) dx$$ Of course, when $k$ is real-valued this is trivial. I'm really just interested in the case where $k$ is purely imaginary. I believe this would follow automatically if I could show that $$\lim_{x\to\infty} F(x) = \lim_{x\to\infty} F(x+k)=L,$$ for all $k \in \mathbb{C}$. The left-hand side limit tells me that for all $\varepsilon >0$, there exists a $\delta >0$ such that $|x| > \delta \implies |F(x)-L| < \varepsilon$. At this point I'd like to say that we can swap out our absolute value signs for complex moduli and claim that $|x+k| > \delta \implies |F(x+k)-L| < \varepsilon$, where this is no longer the absolute value, but the complex modulus. While this sort of seems like it should follow, I can't rigorously justify claiming that something being true under the absolute value with real-valued entries implies that it's also true under the complex modulus and complex-valued entries. After all, we know that $|\sin(x)| \leq 1$ for all $x$, but $|\sin(z)|$ is unbounded.