Another Motivational
Argument for the Expression: eiq= cos q+
i sin q
Problem: Show that eiq
= 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
eix, 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
eu(q)
= eln(cos q + i sin q) = cos q + i sin q.
Our problem thus reduces
to showing that u(q) = iq .We notice immediately that
eu(0)
= cos 0 + i sin 0 = 1
which gives us the identity
u(0) = 1.
We now differentiate the
expression eu = cos q + i sin q to obtain
deu
= eu du = (- sin q + i cos q) dq
or
eu = (- 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.