By Roland Deaux
Aimed at readers unusual with complex numbers, this article explains the way to clear up the categories of difficulties that regularly come up within the technologies, in particular electric reports. to guarantee a simple and entire realizing, themes are built from the start, with emphasis on structures with regards to algebraic operations. 1956 variation.
Read Online or Download Introduction to the Geometry of Complex Numbers (Dover Books on Mathematics) PDF
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Extra resources for Introduction to the Geometry of Complex Numbers (Dover Books on Mathematics)
The points lie on the circle y which they determine. Orient ala, a 2a , au from Za toward Zl, from Za toward Z2' and from Z4 toward Zl; then choose the positive sense of a 24 so that in equation (2), which holds by hypothesis, the integer n is even. The equation then becomes and proves that the point Z4 is on the circle y. Points Za, Z4 mayor may not belong to a common arc defined by the points Zl> Z2' If M is the midpoint of the arc which does not contain Za, the lines MZ a, MZ 4 are bisectors FIG.
The images Z" Z. of the roots of the equation harmonically separate Q" Q. and M is. the midpoint of Z,Z.. See article 31 and exercise 17. [Write the equation as 50 ANHARMONIC RATIO Another procedure, generally less simple. ] Deduce a construction of the images of the roots of the quartic equation Z4 - pz' +q= O. 19. Two points Zl' Z. ] 20. The images Zl' Z. of the roots of the equation az' + 2bz Z~ of the roots of harmonically separate the images Z;, a'z' + + 2b'z + c = 0 c' = 0 if we have ac' - 2bb' + ca' = O.
I ZzW = ZZZ3 ' Z3W = Z3ZZ ' I Z3W ZzW ! 01 = Z3 Z1 ZZZI I and similar equalities for W', whence W, W' are common to the three circles of Apollonius. Z ZZ3 1 = I ZZW,Z 3Z I I = ! ZIZzl and conversely. 20 In order that a triangle ZlZZZ3 be equilateral, it is necessary and sufficient that one of the isodynamic centers be the point at infinity in the Gauss plane,. the other isodynamic center is the point of concurrency of the medians of the triangle. 17 EXERCISES The condition is necessary. have TO 49 31 In fact, if ZlZ2Za is equilateral, we I ZlZa I = I ZzZa I, (21) and consequently (25, 27) (ZlZzZaoo) = (22) e±i3r/S, from which it follows that the point at infinity is an isodynamic center.