Fix an algebraic number $\beta$ and consider a complex number $\alpha$ which admits multiple representations in base $\beta$. If one representation of $\alpha$ is ultimately periodic, must every other representation of $\alpha$ be ultimately periodic?

Bonus question: Does this depend on the choice of the set of digits?

Edit: John Bentin showed that this is false in general, even with $\beta \in \Bbb{Q}$. Is this true at least for $\beta$ algebraic integer?

Edit 2: This question came up when trying to solve this problem of mine.



All I know so far is that if $\beta$ is a Pisot integer, i.e. if it is a real algebraic integer greater than $1$ with every conjugate lying in the unit circle, then $\alpha \in \Bbb{Q}(\beta)$ if and only if it has an ultimately periodic expansion (which is stronger than my question).


Question: Recorded video Frame using AVCaptureSession
Answer:  if ([self.captureSession canSetSessionPreset:AVCaptureSessionPreset320*568]) 
    {
        [self.captureSession setSessionPreset:AVCaptureSessionPreset320*568];
    }


try this one.