By Roland Deaux
Aimed toward readers strange with complex numbers, this article explains the way to clear up the types of difficulties that regularly come up within the technologies, specially electric reports. to guarantee a simple and whole knowing, subject matters are constructed from the start, with emphasis on structures regarding algebraic operations. 1956 version.
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Extra info for Introduction to the Geometry of Complex Numbers (Dover Books on Mathematics)
16 as a function of geometric elements, place arbitrary axes a 13 , au, a 23 , a 24 on the lines ZIZs, Z IZ4' Z2Z3, ZZZ4' which mayor may not be distinct, and designate by Z IZ3, for example, the algebraic value calculated on a l3 of the segment having Zl for initial point and Z3 for terminal point. Assuming (xy) = TT/2, we have (17) + If we should choose a 13 , a 14 , a23 , a 24 so that ZI Z3. R. and an argument is (a 23a I3 ) - (a 24 a U )' This is certainly the case if we take 36 ANHARMONIC RATIO 26.
If (5) does hold, that is, if Zl - Za Z2- Za = reiO or (9) is infinite and the point at infinity of the Gauss plane is the only point Z4 which fulfills the requirement. zll] = re iO . By equating the moduli and then the arguments of these two complex numbers we obtain (7) (8) 9 FIG. 19 30. ll Z Z - r ZlZal· 2 a ' this circle has for diameter the segment PQ of the line Z l Z2' where P and Q are defined by and are thus harmonic conjugates with respect to Zl' Z2. In the case where r 1 = I, that is to say, if r = I ZlZa/Z2Za I, circle y is replaced by the perpendicular bisector of the segment Z1Z2.
24. Show that, for any point M, we have (ABCD) = (ABCM) (ABMD). From this show that, in order to change only the sign of an anharmonic ratio (ABCD), it suffices to replace one point in one of the pairs (A,B), (C,D) by its harmonic conjugate with respect to the other pair. Thus, if (AA'CD) = - I, we have (A'BCD) = - (ABCD). 25. Being given three points A, B, C, we construct the harmonic conjugate of each of them with respect to the other two, so that (AA'BC) = - I, (BB'CA) = - I, (CC'AB) = - 1. By using the properties of article 26 and of exercise 24, show that: 1° (ABCA') = 1/2, (ABCB') = 2; 2° the elimination of C gives (ABA'B') = 4 and (AA'BB') = (BB'CC') = (CC'AA') = - 3° from 1° we obtain (ABC'A') = - 1/2, (ABC'B') = - 2, and, by eliminating A, (BB'C'A') = (CC'A'B') = (AA'B'C') = - 1, whence A is the harmonic conjugate of A' with respect to B', C', etc.