Another Motivational Argument for the Expression: e^iq= cos q+ i sin q

Problem: Show that e^iq = cos q+ i sin q, where i = .

Solution: Let z = x + iy be any complex number. We know, from geometry, that

z = x + iy = r(cos q + i sin q).

In the previous article on e^ix, we used the theory of differential equations to establish the required identity. This time, we will use the natural logarithm function ln(z) to establish that same identity.
Let us form the function ln(z):

ln(z) = ln(x + iy)
= ln [r(cos q + i sin q)]
= ln() + ln(cos q + i sin q).

The first term in the third line, ln(), involves the real number , and so will concern us no further here. The second term, ln(cos q + i sin q), involves a complex number whose magnitude is unity.
Let us set

u(q) = ln(cos q + i sin q)

Then

e^u(q) = e^{ln(cos q + i sin q)} = cos q + i sin q.

Our problem thus reduces to showing that u(q) = iq .We notice immediately that

e^u(0) = cos 0 + i sin 0 = 1

which gives us the identity

u(0) = 1.

We now differentiate the expression e^u = cos q + i sin q to obtain

de^u = e^u du = (- sin q + i cos q) dq
or
e^u = (- sin q + i cos q) dq/du = cos q + i sin q.

We now have the derivative dq/du:

dq/du = (cos q + i sin q)/(- sin q + i cos q)
or
du = [(- sin q + i cos q)/(cos q + i sin q)] dq

Multiplying by unity in the form (cos q - i sin q)/(cos q - i sin q) allows us to simplify the right-hand side, giving

du = i dq.

Integrating, we acquire

u = iq + C.

But since we already know that u = 0 when q = 0, we have that the constant C = 0. Therefore

u = iq,

which establishes the required identity.