Show that $\sin z$ maps a specific set onto the upper-half plane. Show that $\sin z$ maps the set $\{z: - \frac {\pi } {2 }<\operatorname{Re}(z) < \frac {\pi } {2 }, 0< \operatorname{Im}(z) \}  $ onto the upper-half plane.
In the real case the range of $\sin z$ is the interval $[-1,1 ]$. But how can I easiest determine the range of $\sin z$ for a given domain? (For instance what had been the range if the domain had been the set $\{z:0<\operatorname{Re}(z), - \frac {\pi } {2 }<\operatorname{Im}(z) < \frac {\pi } {2 } \}  $ ?)
I know the series definition and the one given by the identity $\sin z = \frac {e^{i z}-e^{-i z } } {2i}  $. Don't know which one is  easier to work with.
Thanks in advance!
 A: Hint: using the exp identity, express $\sin (z)$ as a composition of 3  transformations
and find the range at each step..
Let $f(z)=e^{iz} $and $g(z)=z-1/z $.Then $\sin z=g(f(z))/2i$.
f maps the domain onto right halfdisk $\Omega$.
g takes $\Omega$ onto left half-plane.finally dividing by $2 i$,we get to the upper half-plane.
Note that $g(iz)/i=J(z)$ where J is the  Joukowski mapping $J(z)=z+1/z$.
A: We have:
$$\sin(\sigma + it) = \sin\sigma \cosh t + i \cos\sigma \sinh t $$
hence to find $\sigma\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ and $t\in\mathbb{R}^+$ such that
$$\sin(\sigma + it) = a+bi,$$
with $b>0$, we just need to find a solution for:
$$ a = u\sqrt{1+v^2},\qquad b = v\sqrt{1-u^2}$$
with $u\in(-1,1)$ and $v>0$. 
By eliminating $v$ through the second equation, we just need to prove that:
$$f:(-1,1)\to\mathbb{R},\quad f(u)=u\sqrt{1+\frac{b^2}{1-u^2}}$$
takes any real value over its domain. But this is trivial since $f$ is a continuous function and 
$$\lim_{u\to\pm 1}f(u)=\pm\infty$$
for any value of $b$. Since $f$ is increasing, the map between $\{z: - \frac {\pi } {2 }<\operatorname{Re}(z) < \frac {\pi } {2 }, 0< \operatorname{Im}(z) \}$ and $\{\operatorname{Im}(z)>0\}$ given by the sine function is one-to-one.
