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characterization lemma
If L is the i-line with equation f(z) = azz* -αz* - α*z + b = 0, then
(1) for any z ε C, f(z) is real, and
(2) the sides of L are the regions {z : f(z) > 0} and {z : f(z) < 0}.
(3) if L is a circle, the exterior is the set for which f(z) and a have the same sign.
Proof
(1) is trivial since zz* and (αz*+αz*) are real.
(2) Note that f(z) = 0 if and only if f(z) =0, so the given regions cover C-L.
Now we need to split cases, depending upon the nature of L.
(a) L is a line, so f(z) = -αz* - α*z + b.
Now z,w lie on opposite sides of L id and only if the segement zw cuts L.
That is, if and only if there is a t with 0 < t < 1 and δ = tz + (1-t)w on L.
This amounts to f(δ) = 0, i.e. - αδ* - α*δ + b = 0. Substituing for δ, and
rearranging, we get tf(z) + (1-t)f(w) = 0. Since t and 1-t are positive,
this will occur if and only if f(z) and f(w) have opposite signs.
It follows that z, w are on the same side if and only if f(z), f(w) have the same sign.
(b) L is a circle. Then a ≠ 0, and we can rearrange the equation as
f(z) = a(|z - α/a|2 - r2) = 0, where r2 = (|α|2 - ab)/a2,
so r is the radius of L
and α/a its centre.
Then z is outside L if and only if |z-α/a| > r, i.e. f(z) and a have the same sign.
It follows that z, w are on the same side if and only if f(z), f(w) have the same sign.
(3) was proved in the course of proving (2)(b).
the interior-exterior theorem
If L is an i-line and i an inversive transformation, then
i maps the sides of L to the sides of i(L).
Proof
Since I(2) is generated by inversions, it is enough to consider a single inversion.
By earlier remarks, we may choose axes so that this inversion is i(z) = z* or i(z) = 1/z*.
The first case is easy, as "inversion" is then reflection in a line.
For the second suppose that L has equation f(z) = azz* -αz* - α*z + b = 0.
From Example 1, i(L) has equation f*(z) = bzz* -αz* - α*z + a = 0.
To look at the position of i(z) in relation to i(L), we look at f*(1/z*).
This is (provided z ≠ 0) b/zz* -α/z - α*/z* + a = f(z)/zz*, so f*(1/z*) and f(z) have the
same sign, so if z,w lie on the same side of L, 1/z*,1/w* lie on the same side of i(L),
from part (2) of the lemma.
The case z = 0 needs some comment. This arises only when 0 is not on L, since we are
looking at z on one side of L. Then, from the equation, b ≠ 0, so i(L) is a circle.
If z is sufficiently near 0, then 1/z* will be so large that it lies outside i(L).
Thus points on the same side of L as 0, and close to 0 map to points in the exterior,
and 0 maps to ∞, which
is in the exterior.
It follows that i maps the sides of L to those of i(L).
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