How to show that for arbitrary( complex) trigonometric polynomials $P$ and $Q$ holds
$\frac{1}{2\pi} \int_{-\pi}^{\pi} P(t)Q(mt)dt = \frac{1}{2\pi} \int_{-\pi}^{\pi} P(t)dt \frac{1}{2\pi} \int_{-\pi}^{\pi} Q(t)dt$
always when $m \in \mathbb{Z}$ is sufficiently large.
Hint: Use information that
$\frac{1}{2\pi} \int_{-\pi}^{\pi} e^{int}dt = 0\,\,,\,\,when \,\,n\neq 0\,\,\,,\,\, and \,\,\,\frac{1}{2\pi} \int_{-\pi}^{\pi} e^{int}dt = 1\,\,\, when\,\,n=0$ Just some hint. I know that complex trigonometric polynomials is
$\sum_{n=-m}^{m} c_n e^{int}$
but I also know that if $\,f(t)=t\,$ and $\,g(t)=t+1\,$ , then
$\int_{-\pi}^{\pi} f(t)g(t)dt \neq \int_{-\pi}^{\pi} f(t)dt \cdot \int_{-\pi}^{\pi} g(t)dt$
So I doubt that it is true for complex trigonometric polynomial.