Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

APPENDIX H COMPLEX NUMBERS 0 361
43. Using Euler's formula (6) with y= i· we have ei"/ 3 = cos i + isin i = ~ + '!] i.
47. Taker = 1 and n = 3 in De Moivre'sTheorem to get
(l(cos (} + i sin IJ)] 3 = 1 3 (cos 39 + i sin31J)
(cos IJ + isin 0) 3 = cos 31J + i sin30
cos 3 (} + 3(cos 2 fJ)(i sin 0) + 3(cos IJ)(isln 1J? + (i sin IJ) 3 =cos 38 + i s_in3fJ
cos 3 1J + (3 cos 2 f) sin fJ)·i - 3 cos IJ sin 2 ()- (sin 3 fJ)i =cos 31J + i sin 31J
( cos 3 1J - 3 sin 2 f) cos IJ) + (3 sin f) cos 2 f) - sin 3 1J)i = cos 30 + i ~in 3(}
Equating real and imaginary parts gives cos 30 = cos~ fJ - 3 sin 2 () cos IJ and sin 3B = 3 sin IJ cos 2 (} - sin 3 e.
49. F (x) = er:r: = e
F'(x) = (eax cosbx)' + i(e