Mapping of a region under $w:= \frac{1}{z}$ Find and sketch the image of $x+2y>3$ under the mapping $w:=\frac{1}{z}$.
I know how to do this with a line since the reciprocal mapping will take it to a circle through the origin. I however do not know how to do this algebraically (or even geometrically) properly. Can someone show me the way to do it?  
If the inequality was an equality, I could only do it geometrically as well and not algebraically.
I was thinking for the inequality, it's just "an infinite amount of lines with same slope but increasing y intercept", so it will map to circles with maximal radius $\left|\frac{1}{\frac{1}{3}(1+2i)} \frac{1}{2}\right|$, slowly decreasing to zero as we go further away into the region that's defined.
 A: Yes, as you say understanding the image of just the boundary line is easy if you take it as a given that $1/z$ maps lines not through the origin to circles that pass through the origin. Then, since a circle is determined by three points, you know that the image of the boundary is the circle containing the origin, $z = 1/3$, and $z = -2i/3$ (the latter two points come from looking at where $1/z$ sends the intercepts). 
Your description of the image of the entire region as adding up the images of a bunch of lines is essentially the correct way to think about it. If you want to do it algebraically though then notice that if we write 
$$1/z = u + iv$$
with $z = x+iy$ then 
$$u = \frac{x}{x^2 + y^2},\quad v = \frac{-y}{x^2+y^2}.$$ 
Since $1/z$ is its own inverse we then get the relations
$$x = \frac{u}{u^2 + v^2},\quad y = \frac{-v}{u^2+v^2}.$$ 
So, to understand the image of our region
$$x + 2y > 3$$
in the $u,v$ plane we can just plug in to obtain
$$\frac{u}{u^2+v^2} -2\frac{v}{u^2 + v^2} > 3.$$
If you manipulate this inequality and plug through the algebra (complete the square...) then you should find that the image is the interior of a circle. In particular, it will be the interior of the circle that is the image of the boundary.
A: $ w=\frac1z$ is a Möbius transformation,  or linear fractional transformation. ..  These take circles /lines to lines/circles, in the plane  $\mathbb C \cup \infty $.  $x+2y=3$ is a line.  So choose three points on the boundary line to determine which line or circle it is sent to.  How about $3,\frac32i$ and $\infty $.  These go to  $\frac13,-\frac23i$ and $0$.  These will determine the boundary circle.  Finally choose a point not on the line to determine whether the image is inside or outside the circle.  For instance,  $2i $ .  We get $w=-\frac12i $ , which is inside the circle.  So the image is inside the circle. (For the last step i have used the fact that $w=\frac1z $ is analytic  away from  $0$, hence continuous...
