Harmonic Ratio in Complex Domain
A straight line f(t) = (a + tb)/(1 + t), where t is a(n extended) real parameter and a and b are two complex numbers, passes through points A and B - the geometric images of a and b - such that a = f(0) and b = f(∞).
For any four points on such a line defined by four parameter values p, q, r, s, the cross-ratio is obtained easily in terms of the (real) parameter values:
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(f(p),f(q);f(r),f(s)) = (p,q;r,s).
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Let's verify that this is indeed so.
| (1) |
(f(p),f(q);f(r),f(s)) = (f(p) - f(r))/(f(p) - f(s)) : (f(q) - f(r))/(f(q) - f(s)).
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Evaluating a piece at a time,
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| f(p) - f(r) | = (a + pb)/(1 + p) - (a + rb)/(1 + r) |
| | = [(a + pb)(1 + r) - (a + rb)(1 + p)] / (1 + p)(1 + r) |
| | = (a + pb + ra + prb - a - rb - pa - prb) / (1 + p)(1 + r) |
| | = (pb + ra - rb - pa) / (1 + p)(1 + r) |
| | = (p - r)(b - a) / (1 + p)(1 + r), |
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and similarly for the other three differences in (1). Further,
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| (f(p) - f(r))/(f(p) - f(s)) | = (p - r)(b - a) / (1 + p)(1 + r) : (p - s)(b - a) / (1 + p)(1 + s) |
| | = (p - r)(1 + s) / (p - s)(1 + r). |
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Finally, for the cross-ratio, we see that
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| (f(p),f(q);f(r),f(s)) | = (f(p) - f(r))/(f(p) - f(s)) : (f(q) - f(r))/(f(q) - f(s)) |
| | = [(p - r)(1 + s) / (p - s)(1 + r)] : [(q - r)(1 + s) / (q - s)(1 + r)] |
| | = (p - r)(1 + s)(q - s)(1 + r) / (p - s)(1 + r)(q - r)(1 + s) |
| | = (p - r)(q - s) / (p - s)(q - r) |
| | = (p - r)/(p - s) : (q - r)/(q - s), |
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as promised. We fix now points A(a) and B(b) and concentrate on the values p and q that correspond two points P(p) and Q(q) harmonically conjugate with respect to A and B:
So,
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| -1 | = (p - r)/(p - s) : (q - r)/(q - s) |
| | = (p - r)/(q - r) : (p - s)/(q - s) |
| | = (p - 0)/(q - 0) : (p - ∞)/(q - ∞) |
| | = p/q, |
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which implies q = -p! If we define a conjugation function, say, F(P) = Q, then since q = -p is equivalent to p = -q, the repeated application of F returns the original value:
A function with this property is known as involution. We use this to establish two important properties of harmonic conjugation:
Proposition
For q = -p,
- 1/(f(p) - a) + 1/(f(q) - a) = 2/(b - a),
- (f(p) - f(1))(f(q) - f(1)) = (b - a)2/4.
Both are proved by direct verification. For example,
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| f(p) - a | = (a + pb)/(1 + p) - a |
| | = (a + pb - a - pa)/(1 + p) |
| | = p(b - a)/(1 + p). |
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Similarly
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| f(q) - a | = q(b - a)/(1 + q) |
| | = -p(b - a)/(1 - p). |
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Adding the two terms in #1, we obtain
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| 1/(f(p) - a) + 1/(f(q) - a) | = (1 + p)/[p(b - a)] - (1 - p)/[p(b - a)] |
| | = 2/(b - a). |
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The second part of the proposition is as straightforward. Just note that f(1) = (a + b)/2 is the geometric image of the midpoint of the segment AB. A noteworthy fact about the proposition is that the right hand sides in both identities are independent of the selection of the pair of conjugates (f(p), f(-p)). The second one shows that the points P = f(p) and Q = f(-p) are obtained from each other by inversion in the circle with diameter AB. They lie on the extended diameter and satisfy |OP|·|OQ| = R2, where R = |b - a|/2, the radius of the circle, and O which represents f(1) = (a + b)/2, its center. OP and OQ are signed segments. Assuming the same for AP, AQ, and AB, the first identity reads
which we use in one of the proofs of the Butterfly theorem.
References
- C. Zwikker, The Advanced Geometry of Plane Curves and Their Applications, Dover, 2005
Poles and Polars
Copyright © 1996-2008 Alexander Bogomolny
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