Finding the image of a mapping over a region. I'm having a very hard time understanding the concept of images and mappings in the complex plane.
Considering the map $w=e^{z}=e^{x}e^{iy}$, find the image of the region $\left\lbrace x+iy:x\geq 0, 0\leq y \leq\pi \right\rbrace$. Based on my current understanding, I have rewritten $w=e^z$ by breaking it apart with Euler's Formula:
$$w=e^{x}\left(\cos{y}+i\sin{y}\right)=e^{x}\cos{y}+ie^{x}\sin{y}.$$
From here, we know that $u(x,y)=e^{x}\cos{y}$ and $v(x,y)=e^x\sin{y}$. Could I then rewrite the mapping as $f(x,y)=\left(e^{x}\cos{y},e^x\sin{y}\right)$ in order to sketch the seperate $xy$ and $uv$ planes?
 A: Hint: think of $e^x (\cos y + i \sin y)$ in polar coordinates. 
A: You've written $e^x(\cos y + i\sin y)$.  Now notice that $e^x$ is real, and positive, and $\cos y +i\sin y$ is on the unit circle centered at $0$.  And since $y$ is between $0$ and $\pi$, it's on the top half of the unit circle.  So you've got a positive number times a number on the top half of the unit circle.  The aforementioned positive number can move that point on the top of the circle further from the origin or closer to it.  It simply tells how from from the origin it is.  The precise location on the circle tells you in what direction from the origin it is.
Now notice that that positive number could be any positive number, by choosing $x$ as needed, and that point on the circle could be any point on the top half of the circle, by choosing $y$ as needed.  Look at the picture and you'll see the answer to your question.
Later note: Robert Israel reminds me that there was the constraint that $x\ge 0$.  That would imply that $e^x\ge 1$.  Therefore you only get points on and outside of the unit circle.
