How do we define the branch cuts for $\sin^{-1}z = \frac{1}{i} \log(\sqrt{1-z^2} + iz)$ as $(-\infty,-1)$ and $(1,\infty)$? As $\sin^{-1}z$ is a function of complex $\log$, it is multivalued. The branch cuts to make $\log$ single-valued are defined conventionally as $-\pi < Arg(z) \leq \pi$. Why wouldn't this carry over to $\sin^{-1}z$ and why do these cuts change to $(-\infty,-1)$ and $(1,\infty)$?
 A: The $\sin(z)$ is composition the $f(z)=e^{iz}$ and $g(\zeta)=\frac{1}{2i}(\zeta-\frac{1}{\zeta})$ and we need to see where is biholomorphic. First consider $g$ in $\dot{\mathbb{C}}$. We have $g(\zeta_1)=g(\zeta_2)$ if only if
$$0=\zeta_1-\zeta_2+(\frac{1}{\zeta_2}-\frac{1}{\zeta_1})=(\zeta_1-\zeta_2)(1+\frac{1}{\zeta_2\zeta_1})$$
the show that $g$ is injective on region that does not contain element such that $\zeta_2\zeta_1=1$. Set $G_1=\lbrace\zeta\in\mathbb{C}\vert Re\zeta>0\rbrace$. if $\zeta\in G_1$ then $Re(-\frac{1}{\zeta})=-Re(\frac{\bar{\zeta}}{\vert\zeta\vert^2})<0$, $i.e.$ $-\frac{1}{\zeta}\notin G_1$ and therefore $g$ is injective in $G_1$. Set $G_2=\mathbb{C}-\lbrace t\in\mathbb{R}\vert\vert t\vert\geq1\rbrace$ and we show that $G_2=g(G_1)$ an find the inverse of $g\vert_{G_1}$. For $w\in\mathbb{C}$ & $\zeta\in\dot{\mathbb{C}}$ we have 
(1): $g(\zeta)=w$ if only if $(\zeta-iw)^2=1-w^2$
If $w\notin G_2$ then $w\in\mathbb{R}$ and $w^2>0$. If there is an element $\zeta\in\dot{\mathbb{C}}$ such that $g(\zeta)=w$ and if $\zeta-iw=e^{i\vartheta}$ with $\vartheta\in\mathbb{R}$, the for (1) we have $$(\zeta-iw)^2=e^{i\vartheta}\leq0$$
$i.e$, $2\vartheta=(2k+\pi)$ with $k\in\mathbb{Z}$. It follows that $\vartheta=(k+\frac{1}{2})\pi$ and the $\zeta-iw\in i\mathbb{R}$. Has we had to $w\in\mathbb{R}$ then $\zeta\in i\mathbb{R}$. And for $\zeta\in g^{-1}(w)$ arbitrary, $w\notin g(\mathbb{C}- i\mathbb{R})$. That show  $g(\mathbb{C}- i\mathbb{R})\subset G_2$ and particularly $g(G_1)\subset G_2$. Secondly, if $t\in\mathbb{R}-\lbrace 0\rbrace$, then $g(it)=(t+1/t)/2\in\mathbb{R}$ and
$$\vert g(it)\vert=\frac{1}{4}(t^2+2+\frac{1}{t^2})=\frac{1}{4}(4+(t+\frac{1}{t})^2)\geq1$$
Thus, $g(it)\in\mathbb{C}-G_2$. Since $t\in\mathbb{R}-\lbrace 0\rbrace$ was arbitrary that show
(2):  $g^{-1}(G_2)\subset\mathbb{C}-i\mathbb{R}$
The function $1-w^2$ not vanished on $G_2$ (which is simply connected). Thus exist a logarithm holomorphic for $1-w^2$ in $G_2$ and precisely two branch for the root of $1-w^2$.
Set $h(w)=iw+\sqrt{1-w^2}$ on $G_2$ and we have $g(h(w))=w$. Since $h(0)=1\in G_1$ that follows that $h(G_1)\subset G_2$,$i.e.$ $G_2\subset g(G_1)$. Then $g(G_1)=G_2$ and $h=(g\vert_{G_1})^{-1}$. If $G_0=\lbrace z\in\mathbb{C}\vert-\pi/2<Re(z)<\pi/2\rbrace$ then the restriction $f:G_0\rightarrow G_1$ is biyective with inverse $\frac{1}{i}Log$. Then $\sin\vert_{G_0}=g\circ f$ is biyective whit inverse $$\arcsin(w)=f^{-1}(g(w))=\frac{1}{i}Log(iw+\sqrt{1-w^2})$$
