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Veri cation of the above answers, making use of eq. (153): (a) 5 cos 53:138 j sin 53:138 3 j4; which checks, (b) 5 cos 126:878 j sin 126:878 3 j4; which checks; (c) 5 cos 233:138 j sin 233:138 3 j4; which checks; (d) 5 cos 306:878 j sin 306:878 3 j4; which checks: 93. To nd the SUM of a number of complex numbers we must rst express each one in the form of eq. (156) or (158). We can do this by applying eq. (153) to each term in the given problem; thus 14 cos 1128 j sin 1128 5:2445 j12:9806 8 cos 288 j sin 288 7:0636 j3:7558 19 cos 1558 j sin 1558 17:2199 j8:0298 in which we made use of the identities cos x cos x and sin x sin x. Thus the answer in the form of eq. (156) is 15:4008 j8:7066, which lies in the SECOND QUADRANT of the complex plane, where a 15:4008 and b 8:7066 (see Fig. 110). First, therefore, by eq. (143), A 17:6915. Next we have that h arctan 8:7066=15:4008 29:48108, hence  180 h 150:528, and thus the required answer is (rounded to two decimal places) 17:69 j150:528 94. First apply eq. (161) (extended to cover three factors), then apply eq. (158), thus getting 84 j2158 84 cos 2158 j sin 2158 68:809 j48:180 ; answer: 95. By eq. (164), and also eqs. (157) and (158), we have 12=588 12 cos 588 j sin 588 6:359 j10:177; answer: 96. By eqs. (143) and (144), 2 j3 p j56:30998 13
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Hence, making use of the basic relationship A j n An  jn , we have that p 2 j3 6 13 6  j337:868 2197 cos 337:868 j sin 337:868 2035:01 j827:99; answer: 97. By eq. (163), noting that cos x cos x and sin x sin x,   15 j238 15 cos 238 j sin 238  36 36 0:3835 j0:1628; answer:
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98. One way is to write the problem as the di erence of two fractions, then apply eq. (165), thus 16=1028 9=3908 16 9 =278 =3158 7 7 7=758 7=758 each term now being in the form of eq. (157). Now applying eq. (158) we have 16 9 cos 278 j sin 278 cos 3158 j sin 3158 1:1275 j1:9468; answer: 7 7 99. Setting n 2 in eq. (166) (and writing x in place of  ) you should nd that cos2 x sin2 x j 2 sin x cos x cos 2x j sin 2x hence we have (a) cos 2x cos2 x sin2 x; answer. (b) sin 2x 2 sin x cos x; answer. The above illustrates the fact that the algebra of complex numbers often leads to important and entirely REAL results. 100. Noting that 25 32, and upon setting  488 and n 5 in eq. (166), we have 32 cos 2408 j sin 2408 16:000 j27:713 ; answer: 101. Writing the problem as 0:25 cos 178 j sin 178 3 , then applying eq. (166) for n 3, and remembering that cos x cos x and sin x sin x, we have 0:25 cos 518 j sin 518 0:157 j0:194; answer: 102. First, by eq. (153),  jx  jy cos x j sin x cos y j sin y cos x cos y sin x sin y j sin x cos y cos x sin y By eqs. (161) and (153), or by putting A B 1 in eq. (161), we nd it s also true that  jx  jy  j x y cos x y j sin x y Thus the right-hand sides of the last two equations are equal, and hence, invoking the principle of problem 99, we have (a) cos x y cos x cos y sin x sin y; answer: (b) sin x y sin x cos y cos x sin y; answer: 103. Since 3 j2 lies in the FOURTH QUADRANT of the complex plane, we have (see Fig. 110) h arctan 2=3 33:69018 thus,  360 h 326:30998
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