More "Complex" Complex Analysis

[Pyrokenesis]

I have another problem that has eluded me for days and I'm sure I'm close. If anyone can help, please nudge me in the right direction.

Consider the mapping w = u + iv = 1/z, where z = x + iy. Show that the region between the curves v = -1 and v = 0 maps into the region outside the circle x^2 + (y - 1/2)^2 = 1/4 and above the line y = 0.

I know that w = x/(x^2 + y^2) + i[-y/(x^2 + y^2)].

I also figured since at v = -1, in the x-y plane, y/(x^2 + y^2) = 1.

Alas I know not where to go from here. I have attatched a sketch of the region in the v-u plane.

Thanks in advance.

PS: the question then goes on to ask :- What is the image in the x-y plane of the line -1/2? This is obviously a circle since a straight line in one plane is a curve in the other, but how do I prove this?

[mathwonk]

This mapping is just multiplicative inversion, so before worrying about where it maps any particular region, just understand how it behaves. Since the absolute value of an inverse is the inverse of the absolute values, this maps the points inside the unit circle to points outside the unit circle. E.g. a point located at distance 1/3 from the origin would go to a point at distance 3 from the origin.

But it does not preserve angles. I.e. angles add when you multiply complex numbers, and the number 1 has angle zero, so an inverse has minus the angle of the original number.

So this operation also reflects in the x axis. In fact, inversion just reflects in the x axis, and then inverts radially outward with distance r going to distance 1/r.

first try to see this, and then try your problem.