Compute the principal logarithm of a complex number $z=\sqrt{3}+i$ using $\mathrm{Arg}(z) \in [0,2\pi)$ and $\mathrm{Arg}(z) \in (-\pi,\pi]$.
Wikipedia shows how the answer can be different for the example $z=-i$. On the first interval, the relevant angle is $\frac{3\pi}{2}$ so your answer is $\mathrm{Log}(-i)=\ln(1)+i \frac{3\pi}{2}=i \frac{3\pi}{2}$. On the second interval the relevant angle is $-\frac{\pi}{2}$, so your answer is $\mathrm{Log}(-i)=\ln(1)-i \frac{\pi}{2}=-i \frac{\pi}{2}$.
However for the complex number $z=\sqrt{3}+i$, the relevant angle seems to be the same for both intervals, namely $\frac{\pi}{6}$. So my answer in both cases would be \begin{align*} \mathrm{Log}(\sqrt{3}+i) &=\ln \left(|\sqrt{3}+i|\right) + i \mathrm{Arg} (\sqrt{3}+i) \\ &=\ln \left(2\right) + i \frac{\pi}{6} \end{align*} Am I missing something?
No, your argument (pun intended) is correct. The complex numbers with different logarithm using those intervals would be the ones with negative imaginary part.