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:

(f(p), f(q); f(r),f(s)) = (p, q; r, s).

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)).

Evaluating a piece at a time,

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),

(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).

Finally, for the cross-ratio, we see that

(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),

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:

(p, q; 0, ∞) = -1.

So,

-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,

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:

F(F(P))	= F(Q)
	= P.

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,

f(p) - a	= (a + pb)/(1 + p) - a
	= (a + pb - a - pa)/(1 + p)
	= p(b - a)/(1 + p).

Similarly

f(q) - a	= q(b - a)/(1 + q)
	= -p(b - a)/(1 - p).

Adding the two terms in #1, we obtain

1/(f(p) - a) + 1/(f(q) - a)	= (1 + p)/[p(b - a)] - (1 - p)/[p(b - a)]
	= 2/(b - a).

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

1/AP + 1/AQ = 2/AB,

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

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