By Kiefer N.M., Neumann G.R.
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We already know that to simplify ; : ~: we first obtain a denominator that is entirely real by multiplying top and bottom by ...................... e. 3 - j4 DDDDOODOOOOOOOOOODOOODDDDDDDDDODDDDDDD Right. [(cos e. cos 82 +sin e. sin 82) + j(sin e. cos 82- cos e. sin 82)] r2 1 = 71 [cos ( 81 - 8 2) + j sin(8 1 ,2 - 8 2)] So, for division, the rule is .......................... 18 divide the r's and subtract the angle DOODOODDDDDDDDDODDDDDDDDDDDODDDDDDDDDD That is correct. g. o . o 6(cos 72° + j sin 72°) _ 2(cos 41 o + j sin 41 o)- 3(cos 31 + JSlll 31 ) So we now have two important rules Ifzt =rt(cosOt +j sin8t)andz 2 =r2 (cos8 2 +j sin8 2) then (i) ZtZ 2 = '•'2 [cos(8.
And .......... 6 ODDDDDDODDODDDODDDDDDDDDDDDODODDDDDDOD Now let us consider the following example. First the diagram. Express z = 4- j3 in polar form. From this, y r=5 tan E =~ =0-75 :. E = 36°52' 8 = 360°- 36°52' = 323°81 v, z = 4- j3 = 5( cos 323°81 + j sin 323°8') or in shortened form, z = ................... ,-- --- x - j Y1 But the direction of the vector, measured from OX, could be given as -36°52', the minus sign showing that we are measuring the angle in the opposite sense from the usual positive direction.
4 X 3 X 2) [cos(20° + 30° + 40°) + j sin(20° + 30° + 40°)] = 24( cos 90° + j sin 90°) DDDDODDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD Now what about a few revision examples on the work we have done so far? Turn to the next frame. 46 Programme 2 21 Revision Exercise Work all these questions and then turn on to frame 22 and check your results. 1. Express in polar form, z =-4 + j2. 2. Express in true polar form, z =5( cos 55°-j sin 55°) 3. ) 10(cos 126° + j sin 126°) 11 2(cos72° +jsin72°) 4. Express in the form a+ jb, (i) 2( cos 30° + j sin 30°) (ii) 5( cos 57° - j sin 57°) Solutions are on frame 22.
Search models and applied labor economics by Kiefer N.M., Neumann G.R.