I am trying to find the period of $f(t):+Ae^{-i\lambda t}+Be^{i\lambda t}$ on $\mathbb R$. ($\lambda\in \mathbb R, A, B\in \mathbb C$

To find the period, I guess it suffices to find the period of $g(t):=Ae^{-i t}+Be^{ it}$ and we can find the period of $f$ by rescaling on $t$.

let $L\in \mathbb R$ and $Ae^{-i\lambda ( t+L)}+Be^{i\lambda (t+L)}$=$Ae^{-i\lambda t}+Be^{i\lambda t}$ for all $t$.

After computation, I got $B(e^{iL})^2-(A+B)e^{iL}+A=0$ by setting $t=0$.

Therefore, $e^{iL}=\frac{A}{B}$ or $1$ by applying the formula for quadratic equation. And we can have $L=2n\pi$ or $-iLog(\frac{A}{B})$.

I think $-iLog(\frac{A}{B})$ should be rejected and the period should be $2n\pi$. By assumption, since $g$ is on $\mathbb R$, $L$ should be also a real number, and I guess the imaginary part of $-iLog(\frac{A}{B})$ is most likely non-zero. But I am stuck here.