Final Remarks on Inversion: Alternative Definition of Inversion in Terms of Complex Numbers

The points in the usual coordinate plane $P$ can be thought of as complex numbers: the point $A=(a,b)$ can be thought of as the complex number $z=a+bi$ with $a,b\in\mathbb{R}$. Thus, the $x$-coordinate of $A$ corresponds to the real part of $z$: $\mathbb{R}{\rm e}(z)=a$, and the $y$-coordinate of $A$ corresponds to the imaginary part of $z$: $\mathbb{I}{\rm m}(z)=b$. Recall how we add and subtract complex numbers: this corresponds exactly to addition and subtraction of vectors originating at (0,0) in the plane. For instance, if $z_1=a_1+b_1i$, then $z+z_1=(a+a_1)+(b+b_1)i$; this corresponds exactly to what would happen if we add two vectors $\vec{v}$ and $\vec{v}_1$ which start at the origin and end in $(a,b)$ and $(a_1,b_1)$, respectively: $\vec{v}+\vec{v}_1$ would start at the origin and end in $(a+a_1,b+b_1)$ (cf. Fig. 12.)

Multiplication of complex numbers can be also translated in terms of vectors in the plane. To multiply $z$ and $z_1$ from above, we perform the usual algebraic manipulations:

$$z\cdot z_1=(a+bi)\cdot (a_1+b_1i)=aa_1+ab_1i+ba_1i+bb_1(i^2)= (aa_1-bb_1)+(ab_1+ba_1)i.$$

The resulting ``vector'' $\vec{v}^{\prime}$ from this multiplication corresponds to $(aa_1-bb_1,ab_1+ba_1)$, and it can be interpreted geometrically from the starting vectors $\vec{v}$ and $\vec{v}_1$. I urge you to check in a few simple examples that $\vec{v}^{\prime}$ can be described as follows: add the angles that $\vec{v}$ and $\vec{v}_1$ form with the $x$-axis - this is going to be direction of $\vec{v}^{\prime}$; for the length of $\vec{v}^{\prime}$, take the product of the lengths of $\vec{v}$ and $\vec{v}_1$. (Hint: use the so-called ``polar form'' of vectors and some simple trigonometric identities.)

Question 1. What does this have to do with Inversion?

The function Inversion from the plane $P$ to $P$, as we defined it earlier, can be viewed simply as a complex function, i.e. a function whose input and output are complex numbers. To explain this, we need to introduce one further notion: the conjugate of a complex number. If $z=a+bi$ is a complex number, then the conjugate of $z$, denoted by $\overline{z}$, is simply the complex number obtained from be $z$ by switching the sign of $z$'s imaginary part: $\overline{z}=a-bi$. Geometrically, the points $(a,b)$ and $(a,-b)$ are reflections of each other across the $x$-axis (cf. Fig. 13.) The ``miraculous'' property of conjugates is that their product is always a real number:

$$z\cdot \overline{z}=(a+bi)\cdot (a-bi)=a^2+b^2\in \mathbb{R}.$$

Now we are ready to define Inversion in terms of complex numbers:

Lemma 3   The function Inversion $I:P\rightarrow P$, with center $O=(0,0)$ and radius $r=1$, can be described alternatively by identifying the coordinate plane $P$ with the plane of complex numbers $\mathbb{C}$, and defining the image of $A=(a,b)$ to be the complex number:

$$I(A)=\frac{1}{\overline{z}},$$

where $z=a+bi\in \mathbb{C}$ is the complex number corresponding to $A$.

In other words, Inversion sends the ``point'' $z=a+ib$ to the ``point'' $${\frac{1}{\overline{z}}}$$. The latter has some coordinates produced by the division of the numbers $1$ and $\overline{z}$. Of course, you can say - but how can we divide two complex numbers and get a third complex complex number? Here is an example of how this is done:

$$\frac{1-3i}{2+7i}=\frac{(1-3i)(2-7i)}{(2+7i)(2-7i)}= \frac{-19-15i}{4+49}=-\frac{19}{53}-\frac{15}{53}i.$$

Here we multiplied the numerator and denominator of the original fraction by $(2-7i)$, (the conjugate of $2+7i$), which forced the denominator to become a real number ($53$), and as a result we ended up with an ``ordinary'' complex number.

Thus, according to the lemma, to find where Inversion sends the point $A=(1,1)$, we consider the complex number $z=1+i$, and find the corresponding complex number $1/\overline{z}$:

$$\frac{1}{\overline{z}}=\frac{1}{1-i}=\frac{1+i}{(1-i)(1+i)}= \frac{1+i}{2}=\frac{1}{2}+\frac{1}{2}i.$$

Thus, $A=(1,1)$ will be sent by the Inversion to the point $A_1=(\frac{1}{2},\frac{1}{2})$. Well, it is easy to check that $A_1$ will be indeed the image of $A$ under Inversion: note that $A_1$ lies on the segment $OA$, and $\vert OA\vert\cdot \vert OA_1\vert=\sqrt{2}\sqrt{1/2}=1$. We urge the reader to prove the above lemma by using the elementary properties of complex numbers above and the original definition of Inversion.

Question 2. How good is this new interpretation of Inversion? The original definition seems quite alright, and besides, it does not require knowing complex numbers at all?!

Consider how many cases we have to go through in order to see what happens to circles and lines under Inversion: 4 cases. In addition, the proof of ``preservation of angles'' under Inversion requires us to look at all possible pairs of cases above, making it quite an unattractive work to sweat over ... 10 cases! Besides, the proof in each case has little or no relevance to the other cases, that is, we cannot find one general explanation for why angles should be preserved under Inversion! And honestly speaking, going through all proofs in 10 cases does not really ``impart on us more wisdom'': it only produces technical explanations; we have now no better idea of why Inversion has its wonderful properties than before we started!

In search of a better unifying explanation of why Inversion can do all the miraculous things it does, we invoke the theory of complex functions.

Thus, we consider complex functions $f:{\mathbb{C}}\rightarrow \mathbb{C}$, that is, functions with complex numbers as input and output. For example, $f(z)=z$, $f(z)=3z^2$, $f(z)=\overline{z}$, $f(a+ib)=a+2abi$ are all complex functions. We can also look at functions $f$ defined not on the whole complex plane $\mathbb{C}$, but just on some nice subset of it. For example, $f(z)=1/z$ for $z\not = 0$, and $f(z)=1/\overline{z}$, for $z\not = 0$.

As with real functions (e.g. $f:\mathbb{R}\rightarrow \mathbb{R}$ $f(x)=x^2-4x$,) we can define differentiability of complex functions. We say that a function $f:U\rightarrow \mathbb{C}$, where $U$ is an open subset of $\mathbb{C}$, is complex differentiable at $z_0\in U$ if the limit

$$\lim_{h\rightarrow 0}\frac{f(z+h)-f(z)}{h}$$

exists. We denote this limit, as usual, by $f^{\prime}(z)$. In order not to confuse this definition with the real differentiability, we call a complex differentiable function $f$ holomorphic.

So far so good, except that it is not so obvious when a complex function is holomorphic. We can though describe a whole class of obviously holomorphic functions: these will be polynomials and rational functions of $z$, e.g. $f(z)=z, f(z)=z+3z^2, f(z)=1/z$, but not $f(z)=f(a+bi)=a+2abi$. I shall not elaborate here more on the subject, but just point out a good reference: Complex Analysis, by Serge Lang, Springer-Verlag.

In any case, the story goes roughly as follows.

Theorem 1   Any holomorphic function preserves angles.

More precisely, given two paths in the plane (cf. Fig. 14) meeting at point $A=(a,b)$, we assume that the tangent lines $t_1$ and $t_2$ at $A$ to both paths exist. Set $\alpha$ to be the angle between $t_1$ and $t_2$. After applying a holomorphic function $f$, we transform the two paths into some other paths $f(path\,1)$ and $f(path\,2)$, and they meet at point $B=f(z_0)$. Set $\alpha$ to be the angle between $t_1$ and $t_2$. After applying a holomorphic function $f$, we transform the two paths into some other paths $f(path\,1)$ and $f(path\,2)$, and they meet at point $B=f(z_0)$. Then, the theorem asserts that the new paths will also have tangent lines at $B$, which will make precisely the same angle $\alpha$ with each other. In other words, the angle between the original paths is preserved.

Now, Inversion is not quite a holomorphic function (if it were $f(z)=1/z$ it would have been holomophic everywhere except for $z=0$, where it is not defined anyway.) But inversion $f(z)=1/\overline{z}$ belongs to a class of functions, called, ``antiholomorphic'': roughly speaking, these are functions ``holomorphic'' in the variable $\overline{z}$, not in $z$. Such functions reverse the angles between paths.² As far as the measure of the angles is concerned, it is always preserved under both holomorphic and antiholomorphic functions.

Thus, if the truth, only the truth and the whole truth is to be told,

Theorem 2   Inversion in the plane reverses the angles between any two figures (paths) (as long as we can define such angles.)

Problem 41   Use the formula $f(z)=r^2/\overline{z}$ to describe directly the images of circles and lines passing through (or not through) the center of inversion.