Download Complex Exponential Function: Properties, Applications, and Derivatives and more Study notes Analytical Geometry and Calculus in PDF only on Docsity!
The complex exponential function
1 Comment
You will not need this material in Math 231, but you will need it in later course in mathematics, physics and electrical engineering.
2 Why a complex exponential function should exist
Recall that by definition,
exp(r) = lim n→∞
( 1 + r n
)n
where r is any real number. It will be shown in Math 232 that
exp(r) = lim n→∞
( 1 + r + r^2 2!
r^3 3!
rn n!
) .
(We could do this now with the binomial theorem, but this topic is not a part of this course.) In either case it is the case that we could consider r to be a complex number and have an infinite sequence of complex numbers. In the case where r = it where t is a real number and i^2 = −1 we would get
exp(it) = (^) nlim→∞
( 1 − t^2 2!
t^4 4!
- · · · + (−1)n^ t^2 n (2n)!
)
( t −
t^3 3!
t^5 5!
t^2 n−^1 (2n − 1)!
)
= f (t) + ig(t).
The existence of the two limits, which I have called f (t) and g(t) is easily established via the theorem on limits of monotone functions, and we can show that
f (0) = 1 g(0) = 0 f ′(t) = −g(t) g′(t) = f (t)
all of which allows us to conclude that f (t) = cos(t) and g(t) = sin(t), as pretty remarkable result:
Theorem 1 For each real number t,
exp(it) := limn→∞
( 1 + it n
)n = cos(t) + i sin(t).
For example, exp(2πi) = 1.
It follows from the addition formulae for sine and cosine that for any two real numbers t and s that
exp(it + is) = cos(s + t) + i sin(s + t) = (cos(t) cos(s) − sin(t) sin(s)) + i (sin(t) cos(s) + sin(s) cos(t)) = (cos(t) + i sin(t)) × (cos(s) + i sin(s)) = exp(it) exp(is)
so the exponential function property is still valid. In fact, for any complex number a + bi, we may define exp(a + bi) = exp(a) exp(bi)
and we get an exponential function defined on all complex numbers. By this we mean that exp(x) exp(y) = exp(x + y) for x and y complex numbers, not just real numbers.
3 Applications to trigonometric identities
We have for any real numbers A and B:
(cos(A) cos(B) − sin(A) sin(B)) + i (sin(A) cos(B) + sin(B) cos(A)) = (cos(A) + i sin(A)) (cos(B) + i sin(B)) = exp(iA) exp(iB) = exp((A + B)i) = (cos(A) cos(B) − sin(A) sin(B)) + i (sin(A) cos(B) + sin(B) cos(A)) = cos(A + B) + i sin(A + B)
so the identity exp(iA) exp(iB) = exp(i(A + B) encapsulates both the sine and cosine addition formulae. In fact, it follows by induction that
(cos(A) + i sin(A))N^ = cos(N A) + i sin(N A)
for any real number A and any integer (even negative integers!) N. For example, if we want to find the triple angle formulae for sine and cosine:
cos(3A) + i sin(3A) = (cos(A) + i sin(A))^3 = cos^3 (A) + 3i sin(A) cos^2 (A) − 3 sin^2 (A) cos(A) − i sin^3 (A)
so
cos(3A) = cos^3 (A) − 3 sin^2 (A) cos(A) sin(3A) = 3 sin(A) cos^2 (A) − sin^3 (A)
Observe that we also have
cos(A) = exp(iA) + exp(−iA) 2 sin(A) = exp(iA) − exp(−iA) 2 i
so N∑ − 1
k=
cos(kx) = 1 − cos(N x) − cos(x) + cos((N − 1)x) 2(1 − cos(x))
=
cos(x) − cos((N − 1)x) 4 sin^2 (x/2) N∑ − 1
k=
sin(kx) = − sin(N x) + sin(x) + sin((N − 1)x) 2(1 − cos(x))
= − sin(N x) + sin(x) + sin((N − 1)x) 4 sin^2 (x/2)
Note that these identities could also be derived from (1) and (2).
4 The relation to the geometric properties of complex
numbers
Recall that we may interpret a + bi as a point in the place corresponding to the point (a, b). From the Pythagorean Theorem and the definition of absolute value as the distance from a number to 0 we see that
|a + bi|^2 = a^2 + b^2 = (a + bi)(a − bi)
so |a + bi| =
a^2 + b^2 = |a − bi|. Recall that the number a − bi is called the complex conjugate of a − bi. If |a + bi| 6 = 0 then
a + bi = |a + bi|
( a |a + bi|
) = |a + bi| (cos(θ) + i sin(θ))
where θ is an angle measured (in radians, please) from the ray joining 0 to 1 to the ray joining 0 and a + bi. We usually choose 0 ≤ θ < 2 π, but we don’t have to. Once you choose θ you can replace it by θ + 2nπ where n is any integer. Now that we have the complex exponential function, we see that we can write any non-zero complex number a + bi as exp(c + iθ) where θ is as above and c = ln(|a + bi|). For example,
1 + i =
( √^1 2
√^1
) = exp
( ln(
) .
5 Derivatives
If c is any complex number it is easy to check that d dx exp(cx) = c exp(cx)
by proceeding in two steps. First, write c = a + bi so exp(cx) = exp(ax) exp(ibx). If we can differentiate the second term then we can apply the product rule. d dx
exp(ibx) = d dx
(cos(bx) + i sin(bx)) = −b sin(bx) + ib cos(bx) = ib(i sin(bx) + cos(bx)) = ib exp(bx).
Therefore
d dx exp(cx) = d dx (exp(ax) exp(ibx)) = a exp(ax) exp(ibx) + exp(ax)(ib exp(ibx) = (a + bi) exp(ax) exp(ibx) = c exp(cx)
For example, if f (x) = exp((2 + 3i)x) then f ′(x) = (2 + 3i) exp((2 + 3i)x).
