The number-system of algebra : treated theoretically and historically . g numeri-cally the greater whose corresponding point is the moredistant from the null-point. 48. Problem I. Given the points P and P, representinga-\-ih and a-\-ib respectively; required the point represent-ing a + a -f z (6 + h). The point required is P, the intersection of the parallelto OP through P with the parallel to OP through P. For completing the construction indicated by the figure,we have OD = PE = DD, and therefore OD = 0D-\- OD;and similarly PD = PD + PD. Cor. I. To get the point corresponding to a—a-\-i(b — b
The number-system of algebra : treated theoretically and historically . g numeri-cally the greater whose corresponding point is the moredistant from the null-point. 48. Problem I. Given the points P and P, representinga-\-ih and a-\-ib respectively; required the point represent-ing a + a -f z (6 + h). The point required is P, the intersection of the parallelto OP through P with the parallel to OP through P. For completing the construction indicated by the figure,we have OD = PE = DD, and therefore OD = 0D-\- OD;and similarly PD = PD + PD. Cor. I. To get the point corresponding to a—a-\-i(b — b), produce OP to P, making0P= OP, and complete theptt parallelogram OP, OP. CoR. II. The modulus of thesum or difference of two complexnumbers is less than (at greatestequal to) the sum of their modidi. For OP is less than OP -fPP and, therefore, than OP+ OP, unless 0, P, P are inthe same straight line, whenOP =0P-\- OP. Similarly, PP, which is equal to themodulus of the difference of the numbers represented byP and P, is less than, at greatest equal to, OP + Fig. 2. GRAPHICAL REPRESENTATION OF NUMBERS. 47 49. Problem II. Given P and P, representing a -(- iband a -\- ib respectively; required the j^oint representing(a + i6)(a + 2*6). Let a -}- ib = p (cos 0 -\-i sin 0), § 47 and a-\-ib= p{cos 0+ i sin 0); then {a + ib) (a + ib) = pp(gos 6 + 1 sin 6) (cos ^ + i sin ^) = /3/3[ (cos ^ cos ^ — sin 6 sin ^) + i (sin ^ cos ^ + cos 6 sin ^) ]. But cos 6 cos 6 — sin ^ sin 6 = cos(^ + 6)/^ and • sin^cos^-|-cos^sin^ = sin(^ + ^).* Therefore (a+^&)(a^-^6)=PpTcos(^^-^)^-^sin(^+^)];or, T/ie modulus of the p)roduct of two complex numbers isthe x)roduct of their moduli, its argument the sum of theirarguments. The required construction is, therefore, made by drawingthrough 0 a line making an angle 0-\-0 with OX, and lay-ing off on this line the length pp. CoR. I. Similarly the product of n numbers having moduli9i p\ p, •• p^^ respectively, and arguments 0, 0, 6, ... 0^\is t
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