= 360, 1080, 1800, 2520, 3240 x1 = 1cis(36 ) = 0.81 + 0.59i x2 = 1cis(108) = + .95i –0.31 x3 = 1cis(180) = the only real sol. –2 x4 = 1cis(252) = 0.31 .95i – – x5 = 1cis (324) = 0.81 0.59i – The Primary Solution is x1 = 0.81 + 0.59i The equation y5 = -1 is not the same as y = y5 = -1 has 5 solutions but y = only has 1 solution in the same way that y2 = 9 has 2 solutions y = +3 and -3 but y = 9 ½ only has 1 solution y = +3 So that 0.81 + 0.59i which is verified by the graphics calculator. (It does not = -1) _______________________________________________________________ NB The working for powers 0.5, 0.25, and 0.75 follows: ⅓ Consider Y = x0.5 or x ½ When x < 0 there are no real Y values (ie only complex Y values) so if we express the equation as Y = y + iz = x0.5 where z values are the complex part of the Y values, we can proceed as follows: y + iz = x0.5 Square the equation: (y + iz)2 = x1 y2 z2 + 2yzi = x1 – (N.B. We need x to be the subject of the equation) The x values are to be real numbers < 0 so 2yzi = 0 which implies that either z = 0 or y = 0 If z = 0, the equation just becomes the original y2 = x or y = x0.5 If y = 0, the equation becomes (-z)2 = x1 or x = z2 In Autograph, the basic curve for x 0 is entered as: ≥
= 360, 1080, 1800, 2520, 3240 x1 = 1cis(36 ) = 0.81 + 0.59i x2 = 1cis(108) = –0.31 + .95i x3 = 1cis(180) = –2 the only real sol. x4 = 1cis(252) = – 0.31 – .95i x5 = 1cis (324) = 0.81 – 0.59i The Primary Solution is x1 = 0.81 + 0.59i The equation y5 = -1 is not the same as y = y5 = -1 has 5 solutions but y = only has 1 solution in the same way that y2 = 9 has 2 solutions y = +3 and -3 but y = 9 ½ only has 1 solution y = +3 So that 0.81 + 0.59i which is verified by the graphics calculator. (It does not = -1) _______________________________________________________________ NB The working for powers 0.5, 0.25, ⅓ and 0.75 follows: Consider Y = x0.5 or x ½ When x < 0 there are no real Y values (ie only complex Y values) so if we express the equation as Y = y + iz = x0.5 where z values are the complex part of the Y values, we can proceed as follows: y + iz = x0.5 Square the equation: (y + iz)2 = x1 y2 – z2 + 2yzi = x1 (N.B. We need x to be the subject of the equation) The x values are to be real numbers < 0 so 2yzi = 0 which implies that either z = 0 or y = 0 If z = 0, the equation just becomes the original y2 = x or y = x0.5 If y = 0, the equation becomes (-z)2 = x1 or x = z2 In Autograph, the basic curve for x ≥ 0 is entered as:
05.y=x^(-n)(theory).docx