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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.

Q.matin
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2 Answers2

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

Fly by Night
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  • Very nice crisp answer! But I am stuck on the very last sentence on how you got the perpendicular bisector to be the line $\Re (w) = \frac{1}{2}$? – Q.matin Jan 21 '13 at 22:57
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    Geometrically: The set of points equi-distant from two points is the perpendicular bisector.

    Algebraically: We need $|w|=|w-1|$. Let $w=x+iy$ and consider $|x+iy|=|(x-1)+iy|$. We get, after squaring both sides, $x^2+y^2=(x-1)^2+y^2$ and in turn $x^2=(x-1)^2$. Expanding gives $x^2=x^2-2x+1$ and so $1-2x=0$ or $x=1/2.$

     – Fly by Night Jan 21 '13 at 23:09
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    @Q.matin You're more than welcome. Notice that this method works for any Möbius transformation, i.e.

    $$w = \frac{az+b}{cz+d} , , $$

    where $a,b,c,d$ are complex numbers and $c$ and $d$ are not both zero. Just invert to get $z=\ldots$ and work from there.

     – Fly by Night Jan 22 '13 at 18:48
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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 $\mathrm{Re}(z)=\frac12$.

Community
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robjohn
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  • One question, how did you know to parameterize the circle by $z(t)=1+e^{it}$? – Q.matin Jan 22 '13 at 01:36
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    Because $|z-1|=|(1+e^{it})-1|=|e^{it}|=1$ – robjohn Jan 22 '13 at 01:41
  • Sorry to comment on such an old post, but if you can't immediately see that $\frac{1}{2}-\frac{i}{2}\tan\left(\frac{t}{2}\right)$ is the parametrisation of the line $Re(z)=\frac{1}{2}$, is there a way to work this out? –  Aug 02 '18 at 01:08
  • @Bell: the fact that $\operatorname{Re}\left(\frac12-\frac{\color{#C00}i}2\tan\left(\frac t2\right)\right)=\frac12$ seems pretty straightforward. – robjohn Aug 02 '18 at 01:55
  • Yes of course, I misread the answer. Thanks –  Aug 02 '18 at 02:07
  • What about $z=\rho(1+e^{it})$? – Sigur Apr 23 '19 at 12:35
  • @Sigur: what about it? $|z-\rho|=\rho$ is mapped to the line $z=\frac1{2\rho}+\frac{i}{2\rho}\tan\left(\frac t2\right)$. – robjohn Apr 23 '19 at 12:50