The complex version of the Riemann-Lebesgue lemma

Question

I can't prove the complex version of the Riemann-Lebesgue lemma. $$ f(x) \in \mathbf{C} \\ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos nx \, dx, \quad b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin nx \, dx \\ \frac{1}{\pi} \int_{-\pi}^{\pi} {f(x)}^2 dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty} (a_n^2 + b_n^2) $$ When f(x) is real, the Riemann-Lebesgue lemma $$ \lim_{n \to \infty}a_n=0,\quad \lim_{n \to \infty}b_n=0$$ is obvious. When f(x) is complex, $a_n$'s and $b_n$'s are also complex. Is there a possibility that the LHS converges, $a_n$'s and $b_n$'s of large $n$'s don't limit to zero when $m \to \infty$, and $a_n$'s and $b_n$'s of large $n$'s are cancelling? Can we exclude such a possibility?