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Motivational Argument for the Expression: e^{ix} = cos x + i sin xProblem: Show that e^{ix} = cos x + i sin x, where i = Solution: Let us turn to the theory of differential equations. The equation y'' + y = 0 (where the prime notation symbolizes differentiation with respect to x) has a solution of the form y = cos x + sin x. This same differential equation also has a solution of the form y =e^{ix} + e^{-ix}. [This latter form is obtained by assuming the ansatz, y = e^{m}, and using it in the original equation to obtain the condition on m that m^{2} + 1 = 0, from which m = ± i.] We may now equate cos x + sin x = e^{ix} + e^{-ix} (eq. 1) We wish to solve for e^{ix}. Differentiating once and multiplying the result by -i gives (-
i) [(-sin x) +
(cos x)] = e^{ix}
-e^{-ix}
Adding eqs. 1 and 2 gives 2e^{ix}=
(- i)
cos x + ( + i)
sin x (eq. 3) where * = /2 and * = /2. We must now determine values for * and *. We do so by setting x = 0 and observing that e^{i0} = e^{0} = 1. Also, cos 0 = 1 and sin 0 = 0. In this case, eq. 3a reduces to 1 = * - i* Equating the real and imaginary parts immediately gives * = 1, and * = 0. With these values, eq. 3a yields the required expression e^{ix}
= cos x + i sin x Q.E.D. |
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