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Consider the locus C(a,α,b) defined by the equation
azz* - αz* - α*z + b = 0
where, if a = 0, then α ≠ 0, and if a ≠ 0, then |α|2 > ab.
In fact, these amount to the single condition |α|2 > ab.
To get equations in x and y, put z = x+iy, α = f+ig.
If a = 0, then we have a line with xy-equation -2fx - 2gy + b = 0.
If a ≠ 0, then we have a circle with xy-equation a{x2 + y2 -2(f/a)x -2(g/a)y + b/a} = 0.
The condition on α guarantees a circle. The radius is r, where r2 = (|α|2 - ab}/a2, and
the centre is (f/a,g/a). In complex terms, we say the centre is α/a.
It is easy to see that any line or circle has an equation of the given form, and that
the values of a,b and α are unique up to scaling by a real factor.
It is also useful to look at the function F(w,z) = awz* - α*w - αz* + b.
Then the above locus is F(z,z) = 0.
Observe that F(z,w) = azw* - α*z - αw* + b = F(w,z)* as a,b are real.
Thus, F(w,z) = 0 if and only if F(z,w) = 0. This suggests the following
Definition
Complex numbers z,w are inverse with respect to C(a,α,b) if F(w,z) = 0.
With L = C(a,α,b), we associate the map inversion in L defined by
iL(z) = (αz* - b)/(az* - α*), where, if a ≠ 0, then z ≠ α/a.
The map is obtained by solving F(w,z) = 0 to get w as a function of z.
Observe that iL(z) = z if and only if z ε L.
geometrical interpretation
(1) If L is a line then iL is reflection in L.
(2) If L is a circle, centre C, radius r, then iL maps P ≠ C to Q, where
Q lies on the ray from C through P, and |CP||CQ| = r2.
proof
Observe that these descriptions do not depend on the choice of coordinates, or the scale since
they involve only ratios of lengths - the condition in (2) may be written (|CP|)/r)(|CQ|/r) = 1.
This means that, if L is a line then we may choose coordintes so that L is the real axis.
In our notation, this has equation iz* + i*z = 0 (i.e. z* = z). Then iL(z) = iz*/(-i*) = z*.
If L is a circle then we may choose coordintes so that L is the circle |z| = 1.
In our notation, this has equation zz* - 1 = 0. Then iL(z) = 1/z*.
It follows that investigation of inversion reduces to two cases with very simple formulae.
The interpretation also makes it clear that the square of an inversion is the identity on its domain.
In the case of a circle with centre C, this is C-{C}.
We observed that inversion t with respect to an such a locus has order 2.
This makes the calculation of images under t rather easy.
Suppose that a curve L has equation f(z) = 0.
Then z ε t(L) if and only if t(z) ε t2(L) = L.
Thus t(L) has equation f(t(z)) = 0.
Example
Find the inverses of C(a,α,b) with respect to (a) the real axis, and (b) the circle |z| = 1.
Solution
Note that C(a,α,b) has equation f(z) = azz* -α*z - αz* + b = 0.
(a) Earlier, we saw that inversion with respect to the real axis is given by i(z) = z*.
Thus the image has equation f(z*) = az*z -α*z* - αz + b = 0. This is C(a,α*,b).
(b) Inversion with respect to |z| = 1 is given by j(z) = 1/z*.
Thus, the image has equation f(1/z*) = a/zz* -α*/z* - α/z + b = 0 (ignoring z = 0).
Multiplying by zz*, we get a -α*z - αz* + bzz* = 0. The image is C(b,α,a).
Note that we have chosen conditions on a, α and b to ensure that the loci are non-trivial,
i.e. are infinite, but not the entire plane. Even ignoring a point if necessary, the images
will also be non-trivial, so satisfy the conditions. This can be checked by algebra.
In the example we inverted in just two particular cases. But, by earlier remarks, every
such locus has one of the above forms for suitable choice of axes. Thus we have shown
that inversion always maps C(a,α,b) to C(c,γ,d), for some c, γ, d.
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