Continuing the logarithm once around the unit circle adds 2 pi i

For $t\in [0,1]$, let $V_t:=e^{2\pi it}B(1,1\mathrm{/}2)$ and define $L_t(u):=2\pi it+\mathrm{Log}(e^{-2\pi it}u)$ for $u\in V_t$. By [L1], each $L_t$ is holomorphic on $V_t$ and satisfies $\exp (L_t(u))=u$ and $L_t(\gamma (t))=2\pi it$.

Use the subdivision $t_k:=k\mathrm{/}16$ for $0\le k\le 16$. For $t\in [t_k,t_{k+1}]$, $\mathrm{\mid }\gamma (t)-\gamma (t_k)\mathrm{\mid }\le 2\sin (\pi \mathrm{/}16)<\frac{1}{2},$ so the whole subpath lies in $V_{t_k}$. At the joining point one has $L_{t_k}(\gamma (t_{k+1}))=2\pi i{t}_k+\mathrm{Log}(e^{2\pi i\mathrm{/}16})=2\pi i{t}_{k+1}=L_{t_{k+1}}(\gamma (t_{k+1})),$ because the argument increment is $\pi \mathrm{/}8$. Thus the branches form an admissible continuation chain along $\gamma$. At $t=0$ this is the principal logarithm germ, while the terminal germ at $1$ has value $2\pi i$. By [L2], it is the germ of $\mathrm{Log}+2\pi i$ at $1$.