PureMath.pdf

# Now let ob x y oc x y od x y and suppose that a is

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Now let OB = [ x, y ] , OC = [ x 0 , y 0 ] , OD = [ X, Y ] , and suppose that A is the point (1 , 0), so that OA = [1 , 0]. Then OA · OD = [1 , 0][ X, Y ] = [ X, Y ] , and so [ x, y ][ x 0 , y 0 ] = [ X, Y ] . The product OB · OC is therefore to be defined as OD , D being obtained by constructing on OC a triangle similar to OAB . In order to free this

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[III : 38] COMPLEX NUMBERS 91 O A B C D D Fig. 22. definition from ambiguity, it should be observed that on OC we can de- scribe two such triangles, OCD and OCD 0 . We choose that for which the angle COD is equal to AOB in sign as well as in magnitude. We say that the two triangles are then similar in the same sense . If the polar coordinates of B and C are ( ρ, θ ) and ( σ, φ ), so that x = ρ cos θ, y = ρ sin θ, x 0 = σ cos φ, y 0 = σ sin φ, then the polar coordinates of D are evidently ρσ and θ + φ . Hence X = ρσ cos( θ + φ ) = xx 0 - yy 0 , Y = ρσ sin( θ + φ ) = xy 0 + yx 0 . The required definition is therefore [ x, y ][ x 0 , y 0 ] = [ xx 0 - yy 0 , xy 0 + yx 0 ] . (6) We observe (1) that if y = 0, then X = xx 0 , Y = xy 0 , as we desired; (2) that the right-hand side is not altered if we interchange x and x 0 , and y and y 0 , so that [ x, y ][ x 0 , y 0 ] = [ x 0 , y 0 ][ x, y ];
[III : 39] COMPLEX NUMBERS 92 and (3) that { [ x, y ] + [ x 0 , y 0 ] } [ x 00 , y 00 ] = [ x + x 0 , y + y 0 ][ x 00 , y 00 ] = [( x + x 0 ) x 00 - ( y + y 0 ) y 00 , ( x + x 0 ) y 00 + ( y + y 0 ) x 00 ] = [ xx 00 - yy 00 , xy 00 + yx 00 ] + [ x 0 x 00 - y 0 y 00 , x 0 y 00 + y 0 x 00 ] = [ x, y ][ x 00 , y 00 ] + [ x 0 , y 0 ][ x 00 , y 00 ] . Similarly we can verify that all the equations at the end of § 37 are satisfied. Thus the definition (6) fulfils all the requirements which we made of it in § 37 . Example. Show directly from the geometrical definition given above that multiplication of displacements obeys the commutative and distributive laws. [Take the commutative law for example. The product OB · OC is OD ( Fig. 22 ), COD being similar to AOB . To construct the product OC · OB we should have to construct on OB a triangle BOD 1 similar to AOC ; and so what we want to prove is that D and D 1 coincide, or that BOD is similar to AOC . This is an easy piece of elementary geometry.] 39. Complex numbers. Just as to a displacement [ x ] along OX correspond a point ( x ) and a real number x , so to a displacement [ x, y ] in the plane correspond a point ( x, y ) and a pair of real numbers x , y . We shall find it convenient to denote this pair of real numbers x , y by the symbol x + yi. The reason for the choice of this notation will appear later. For the present the reader must regard x + yi as simply another way of writing [ x, y ]. The expression x + yi is called a complex number . We proceed next to define equivalence , addition , and multiplication of complex numbers. To every complex number corresponds a displacement. Two complex numbers are equivalent if the corresponding displacements are equivalent. The sum or product of two complex numbers is the complex

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[III : 39] COMPLEX NUMBERS 93 number which corresponds to the sum or product of the two corresponding displacements. Thus x + yi = x 0 + y 0 i, (1) if and only if x = x 0 , y = y 0 ; ( x + yi ) + ( x 0 + y 0 i ) = ( x + x 0 ) + ( y + y 0 ) i ; (2) ( x + yi )( x 0 + y 0 i ) = xx 0 - yy 0 + ( xy 0 + yx 0 ) i. (3) In particular we have, as special cases of (2) and (3), x + yi = ( x + 0 i ) + (0 + yi ) , ( x + 0 i )( x 0 + y 0 i ) = xx 0 + xy 0 i ; and these equations suggest that there will be no danger of confusion if, when dealing with complex numbers, we write x for x +0 i and yi for 0+ yi , as we shall henceforth.
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