Find the image of the strips $0\le \text{Re}(z) \le$ and $0\le \text{Im}(z) \le 2$ under the map $w=1/z$. Find the images of the strips $0\le \text{Re}(z) \le 2$ and $0\le \text{Im}(z) \le 2$ under the map $w=1/z$. 
I tried using this fact from the text:

Then for the line $\text{Re}(z)=2$ I get the circle $|z-1/4|=1/4$. And the answer says that $0\le \text{Re}(z) \le 2$ maps onto the right half-plane minus the disk $|w-1/4|\lt 1/4$. But I don't see how I can argue this. I'm quite clueless on how to progress with such problems. I would greatly appreciate any help on this.
 A: If you try to do it abstractly in one step, you will fall into the pit fast. It's better to break it into pieces and then use your text's results.
This is the transformation of a circle inversion and the exercise purposefully specifies an overlap of the strip with the unit circle, which are the fixed points of the transformation.
The inversion is holomorphic away from any open neighborhood of infinity, so it pays again to parametrize some obvious open sub-cover counterclockwise, the images of the boundaries of which can be seen to be what your text says they are.
It's fairly useful to avoid looking (close) at the origin, as this goes to infinity and to avoid being close to the unit disk, since the behavior of the transform is different for interior and exterior points.
Using the following notation then,
$$z(\rho,\theta)=\rho\cdot\exp(i\theta)$$
If you parametrize the boundaries of your strip domain, for example, as (for some reason LaTex doesn't render my right braces here, I don't know why):
$$
\begin{align}
s_1&=\{z(\rho,0)\colon 1+\epsilon\lt\rho\le 2\}\\
s_2&=\{z(2,0)+ti\colon 0\le t\le 2\}\\
s_3&=\{t+2i\in\colon 0\le t\le 2\}\\
s_4&=\{z(\rho,\pi/2)\colon 1+\epsilon\lt\rho\le 2\}\\
s_5&=\{z(1+\epsilon,\theta)\colon 0\le\theta\le\pi/2\}\\
s_6&=\{z(1-\epsilon,\theta)\colon 0\le\theta\le\pi/2\}\\
s_7&=\{ti\colon \epsilon\lt t\lt1-\epsilon\}\\
s_8&=\{z(\epsilon,\theta)\colon 0\le \theta\le\pi/2\}\\
s_9&=\{t\colon \epsilon\lt t\lt 1-\epsilon\}\\
\end{align}
$$
Your domain now looks like this:

and now using your text's advice you can individually identify the images of the various segments, as the following: .i.e. the outer segments map into arcs in the Poincare disk, lines to lines and exterior circles into interior circles, with the innermost circle becoming a neighborhood of infinity:

A: The circle $\lvert z - \frac{1}{4} \rvert = \frac{1}{4}$ in the complex plane corresponds to the circle $(x-\frac{1}{4})^2 + y^2 = \big(\frac{1}{4}\big)^2$ in the $xy$-plane, which can be rewritten as $a(x^2+y^2)+bx+cy+d$, where $(a,b,c,d)=(1, -\frac{1}{2}, 0, 0)$. Applying your hint gets you your straight line as desired.
A: The Moebius transformation $f:\ z\mapsto{1\over z}$, supplied with $f(0):=\infty$, $\> f(\infty):=0$, maps
the segment $\> ]0,2]$ onto the infinite segment $\bigl]\infty,{1\over2}\bigr]\subset{\mathbb R}$, and maps the segment $\>]0,2i]$ onto the infinite segment $\bigl]\infty,-{i\over2}\bigr]\subset i{\mathbb R}$.
As $f(2+2i)={1\over4}(1-i)=:\zeta$ the segment $[2, 2+2i]$ is mapped onto an arc of a circle intersecting the real axis at ${1\over2}$ orthogonally and going through $\zeta$. It is then geometrically obvious that this arc is a quarter of a circle of radius ${1\over4}$. Similarly, the segment $[2i, 2+2i]$ is mapped onto an arc of a circle intersecting the imaginary axis at $-{i\over2}$ orthogonally and going through $\zeta$. It is then again obvious that this arc is a quarter of a circle of radius ${1\over4}$. 
As $f(1+i)={1\over2}(1-i)$ it follows that  the square $Q:=\{z=x+iy\in{\mathbb C}\>|\>0\leq x\leq2, \ 0\leq y\leq 2\}$  is mapped onto the fourth quadrant minus the two "buttocks" near the origin.
