# 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?

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Try to find out the image of the vertical lines $\{(x_0,y) : y>0\}$ for each $x_0\in(-\pi/2,\pi/2)$. I remember something nice was obtained. – emiliocba Feb 1 '12 at 13:27

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 $x_0$)

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