Image of a map in the complex plane Is there an elegant way (either intuitive/ by a series of diagrams or by manipulating numbers/algebra) to find out what the image of $\sin(w)$ where $w\in \mathbb C$ from a domain say $\{w\in \mathbb C| Re(w)\in(-\pi/2,\pi/2), Im(w)>0\}$? I tried writing $\sin(w)={e^{iw}-e^{-iw}\over 2i}$ but that isn't too helpful... Maybe there is a clearer way? Actually I know that the answer is the upper half-plane, but how to get there?
 A: The entire function $sin$ maps the three rectilinear pieces of the boundary of your domain $D$ onto respectively the closed intervals $[-\infty,-1]$, $[-1,+1]$ and $[+1,+\infty]$ of the real axis.
This implies  that the boundary of $D$ is mapped onto the real axis.
On the other hand, the positive $y$-axis   is mapped bijectively onto itself (recall that $sin(iy)=i\cdot sh(y))$.
This  strongly suggests that the required image is the upper half plane .
To see it rigorously, the most straightforward way is to brutally compute the image of the rays $Re(z)=Re(x+iy)=x_o= \;$constant of your domain $D$ under the $sine$ function, as suggested by emiliocba.  
If $x_0\gt 0$ for example you will find the upper right quarter of the hyperbola     $$X=sin(x_0)\cdot cosh(y)\quad  Y=cos(x_o)\cdot sinh(y)$$
and these quarters of hyperbolas fill the upper right quarter plane of the image complex plane.
If $x_0=0$, we have already seen that the positive $y$-axis maps onto itself,  and I'll leave to you the case $x_0\lt 0$.
Edit
Of course you can also look at the images of the horizontal segments $(-\pi/2,\pi/2)\times \lbrace y_0\rbrace \subset D$ under the function $sine$.
They cover the upper half-plane, since they are the  upper halves of the ellipses
$$      X=sin(x)\cdot cosh(y_0) \quad Y=cos(x)\cdot sinh(y_0)                $$
(This is even slightly simpler, in that you don't have to consider separate cases as above for $x_0$)
