Inversion mapping complex function

Question

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.

Answer 1

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 $\Re (w) = \frac{1}{2}$.

Answer 2

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 $\mathrm{Re}(z)=\frac12$.