What does e^{ix} mean? Here i^{2} = −1. Recall the complex numbers are of the form a + ib and
are identified as points in the plane. For f

(x)

= e^{ix}, you would want

′′ 2
f (x) = if (x ) = − f (x )

so

f′′(x) + f (x) = 0.

Also, you would want

f (0) = e0 = 1,f′(0) = ie0 = i.

One solution to these conditions is

f (x ) = cos(x) + isin(x).

Is it the only solution? Suppose g

(x)

is another solution. Consider u

(x)

= f

(x)

−g

(x)

. Then
it follows

u′′(x)+ u(x) = 0,u (0) = 0 = u′(0).

Thus both Reu and Imu solve the differential equation and 0 initial condition. By Lemma
8.3.3 both Reu and Imu are equal to 0. Thus the above is the only solution. Recall by
De’Moivre’s theorem