# Harmonic Ratio in Complex Domain

A straight line

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), |

and similarly for the other three differences in (1). Further,

(f(p) - f(r))/(f(p) - f(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)(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(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

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

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

- Poles and Polars
- Brianchon's Theorem
- Complete Quadrilateral
- Harmonic Ratio
- Harmonic Ratio in Complex Domain
- Inversion
- Joachimsthal's Notations
- La Hire's Theorem
- La Hire's Theorem, a Variant
- La Hire's Theorem in Ellipse
- Nobbs' Points, Gergonne Line
- Polar Circle
- Pole and Polar with Respect to a Triangle
- Poles, Polars and Quadrilaterals
- Straight Edge Only Construction of Polar
- Tangents and Diagonals in Cyclic Quadrilateral
- Secant, Tangents and Orthogonality
- Poles, Polars and Orthogonal Circles
- Seven Problems in Equilateral Triangle, Solution to Problem 1

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