Possible error in Complex Analysis with Applications by Asmar: $n\log_{\alpha}(z) = \log_{\beta}(z) + 2k\pi i, z \neq 0$ The author (Asmar) states in the page 88 of Complex Analysis with Applications that:

Since two values of $\mathrm{arg}(z)$ differ by an integer multiple of $2\pi$, it follows that, for a complex number $z \neq 0$ and a real numbers $\alpha, \beta$, there is an integer $k$ (depending on $z, \alpha, \beta$), such that $n\log_{\alpha}(z) = \log_{\beta}(z) + 2k\pi i$

when the $\alpha$-th branch of $\mathrm{log}(z)$ is defined as: $\log_{\alpha}(z) = \ln(|z|) + i \mathrm{arg}_{\alpha}(z), \mathrm{arg}_\alpha(z) \in (\alpha, \alpha + 2\pi]$.
Surely the LHS of the equation should not have the $n$ (presumably $n \in \mathbb{N}$) as a multiple? Since otherwise this would imply that $(n - 1)\ln(|z|) = 0, \forall z \in \mathbb{C}\setminus \{0\}$.
 A: You are right, it is certainly a typo.
Any logarithm branch $\log_\alpha$ has the property $e^{\log_\alpha(z)}= z$ for all $z \in \mathbb C \setminus \{0\}$. Note that $\log_\alpha$ is not continuous in the points lying on the branch cut $B_\alpha = \{re^{i\alpha} \mid r > 0\}$.
Thus we have $e^{\log_\alpha(z)} = e^{\log_\beta(z)}$ which means that $e^{\log_\alpha(z)- \log_\beta(z)} = 1$, i.e. $\log_\alpha(z)- \log_\beta(z) = 2k\pi i$ with some $k \in \mathbb Z$. The number $k$ depends on $\alpha,\beta$ and $z$.
Now there are two cases.

*

*The two branch cuts $B_\alpha$ and $B_\beta$ agree. This means that $\alpha - \beta =  2l\pi i$ with some $l \in \mathbb Z$. Then we have $k = l$ and there is no dependency on $z$.


*The two branch cuts $B_\alpha$ and $B_\beta$ do not agree. In that case  they split $\mathbb C \setminus \{0\}$ into two connected regions $R_j = R_1, R_2$ and on each of them we have $\log_\alpha(z)- \log_\beta(z) = 2k_j\pi i$ with $k_j$ not depending on $z$. However, we have $k_2 = k_1 \pm 1$.
