Inversion mapping complex function 
Show that the inversion mapping $w = f(z) = \frac{1}{z}$ maps: the
  circle $|z-1|=1$ onto the vertical line $x=\frac{1}{2}$.

From what I know thus far, I can see that $|z-1|=1$ take $\theta$ from $2\pi > \theta > 0$ will traverse the circle at $z= 1 + e^{i\theta}$, am I right on that since the graph of the function is shifted to the right with radius one, thus $z= 1 + e^{i\theta}$? I do not know how to finish the proof.
 A: You have a function $w=f(z)$ where $f(z) = \frac{1}{z}$. 
The trick to this types of question is to compute the inverse. If $w = \frac{1}{z}$ then $z=\frac{1}{w}$. It follows:
$$z-1 = \frac{1}{w}-1 = \frac{1-w}{w} \, . $$
Taking the modulus of both sides gives: 
$$|z-1|=\left| \frac{1-w}{w}\right| =\frac{|1-w|}{|w|} \, . $$
We know that $|z-1|=1$ and so, by multiplying both sides by $|w|$, we get $|w|=|1-w|$, or equivalently, $|w|=|w-1|$. By definition, this is the set of points $w$ equi-distant from $w=0$ and $w=1$, i.e. the perpendicular bisector: the line $\Re (w) = \frac{1}{2}$.
A: As shown in this answer, this fact can be proven geometrically.
Alternatively, we can parametrize the circle by
$$
z(t)=1+e^{it}
$$
Then, inverting maps this to
$$
\begin{align}
\frac1{z(t)}
&=\frac1{1+e^{it}}\\
&=\frac1{1+\cos(t)+i\sin(t)}\frac{1+\cos(t)-i\sin(t)}{1+\cos(t)-i\sin(t)}\\[6pt]
&=\frac{1+\cos(t)-i\sin(t)}{(1+\cos(t))^2+\sin^2(t)}\\[6pt]
&=\frac{1+\cos(t)-i\sin(t)}{2+2\cos(t)}\\[6pt]
&=\frac12-\frac i2\tan\left(\frac t2\right)
\end{align}
$$
which parametrizes the line $\mathrm{Re}(z)=\frac12$.
