Let the vector $ |V\rangle = r_0 e^{i\theta_0} |0\rangle + r_1 e^{i\theta_1} |1\rangle $ correspond to the state of a qubit where $r_0,r_1,\theta_0,\theta_1 \in \mathbb{R}$.
According to p. 22 of Quantum Computing: A Gentle Introduction by Rieffel & Polak, the relative phase of $|V\rangle$ in this standard basis is the unit modulus complex number $e^{iX}$ such that
$$ \frac{r_0e^{i\theta_0}}{r_1e^{i\theta_1}} = e^{iX} \frac{|r_0e^{i\theta_0}|}{|r_1e^{i\theta_1}|}. $$
Hence, I derive the relative phase of $|V\rangle$ to be $e^{i(\theta_0-\theta_1)}$ as follows
$$ \frac{r_0e^{i\theta_0}}{r_1e^{i\theta_1}} = e^{iX} \frac{|r_0|}{|r_1|} = e^{iX} \frac{r_0}{r_1} $$
$$ \Leftrightarrow e^{iX} = \frac{r_1}{r_0} \cdot \frac{r_0e^{i\theta_0}}{r_1e^{i\theta_1}} = \frac{e^{i\theta_0}}{e^{i\theta_1}} = e^{i\theta_0-i\theta_1} = e^{i(\theta_0-\theta_1)}. $$
However, I have found a few sources stating the relative phase of $|V\rangle$ is defined to be $e^{i(\theta_1-\theta_0)}$, which is the conjugate of what I derived above using the textbook definition.
Can someone help me make sense of this difference? Is there an error in the textbook definition, or am I making some kind of mistake in my calculation?
As I think about it, I believe either answer would be appropriate depending on which modulus one complex number, either $e^{i\theta_0}$ or $e^{i\theta_1}$, you factor out of $|V\rangle$ to distinguish it from all other vectors equal up to global phase. Could someone confirm? Is there a convention on which phase factor one would factor out?
P.S. I did find a related post here, but it doesn't seem to go into sufficient depth to answer my question. Moreover, the author seems to designate the difference $\theta_1 - \theta_0$ or $\theta_0 - \theta_1$ as a relative phase as opposed to the entire phase factor $e^{i(\theta_1-\theta_0)}$ or $e^{i(\theta_0-\theta_1)}$.