Motivational Argument for the Expression: e^ix = cos x + i sin x

Problem: 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 m² + 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
or
i [sin x - cos x] = e^ix -e^-ix (eq.2)

Adding eqs. 1 and 2 gives

2e^ix= (- i) cos x + ( + i) sin x (eq. 3)
or
e^ix = (* - i*) cos x + (* + i*) sin x (eq. 3a)

where * = /2 and * = /2. We must now determine values for * and *. We do so by setting x = 0 and observing that eⁱ⁰ = e⁰ = 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.