Download Complex Numbers in Geometry by I. M. Yaglom, Henry Booker, D. Allan Bromley, Nicholas PDF

By I. M. Yaglom, Henry Booker, D. Allan Bromley, Nicholas DeClaris

ISBN-10: 1483256634

ISBN-13: 9781483256634

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This condition may be written: ZΛn Ζ Zo Z9 Z(\ ΖΛ £3 Zr> Zn *1 z2 #2 z* (12) §7. Ordinary Complex Numbers as Points of a Plane 33 From this it follows that the equation of the circle (or line) S passing through the points zx , z2 , and z3 has the form % #2 Z^ Z2 . % -^3 #1 Z$ Z Z-^ Z 2 . Z2 % ^3 Zi Z% /i o\ (z - z2){z — ζ3)[{ζ± — z3){zx - z2)] = (*— *3)(* - *2)[(*1 - *2)(*1 — *s)] That is, Azz + Bz - Ëz + C = 0 where A ={z± — z3){z± - z2) — {zx — z2)(z1 — z3) B = -z3{zx C = Z2Z3{Z1 - * 3 )(*! - z2) + z2(z1 - z2){z± ^3/(^1 ^2) ^3^21*1 ^X^l z3) ^3/ Thus, the equation of every circle {or line) may be written in the following form: Azz + Bz — Bz + C = 0, A C purely imaginary (14) Conversely, the locus of points z which satisfy any equation of this form is a circle or a line (if such points exist).

In that case it is obvious that the point ax + a2 = hz is a vertex of a rhombus Oaxh^a2 , and so the lines [Oh3] and [αλα2] are perpendicular (since they are diagonals of the rhombus); the 50 II. Geometrical Interpretation of Complex Numbers point ms = hJ2 = (a1 + ^2)/2 is the midpoint of the side αλα2 of the triangle axa2a3 . Further, the point ( = ns + as) h = ax + a2 + % is a vertex of a parallelogram Ohsha3 . In other words, the line [a3h] || [Oh3] J_ [^^2]; that is, the line [a3h] is an altitude of the triangle axa2a3 , and the point ¿ 3 where it meets the side [a^] is the foot of the altitude.

10 43 §8. Applications and Examples where, as before, the braces emphasize that it is a question of oriented lengths of segments; that is, the product {O, z^ · {O, z2} is counted as positive if the directions of the s e g m e n t s Ozx and Oz2 (from O to ζλ and from O to z2 , respectively) coincide and the points z1 and z2 lie on the same side of O, and negative if the directions of these segments are opposite (the points zx and z2 lie on opposite sides of O). Since the points zx and z2 lie on the circle S> Azxzx +Bz1- Bz± + C = 0 (16) + C = 0 (17) Az2z2 + Bz2-Bz2 On the other hand, since these points lie on a line through the origin O, arg z2 = arg zx and arg z2 = —arg z1 (Figure 10a), or arg z2 = arg ζλ + π and arg z2 = —arg z1 — π (Figure 10b), and consequently the product ζλζ2 is real in all cases: zxz2 = ky (18) ζλζ2 = ζλζ2 = k = k We now multiply Equation 16 by z2 and Equation 17 by zx and use Equations 18; we obtain Akzx + Βζλζ2 — Ëk + Cz2=0 (16a) Akz2 + Βζλζ2 - Ëk + Cz1=0 (17a) Subtracting Equation 17a from Equation 16a, we have Ak{zx - z2) - C{zx -z2)=0 whence it follows that, if zx Φ z2 , so that zx — z2 Φ 0, then h But the product Z-iZn ~ A #v exactly coincides with the product {O, z-^ · {O, z2} of the lengths of the (oriented) segments Oz1 and Oz2 .

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