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Math 307
THE COMPLEX EXPONENTIAL FUNCTION
(These notes assume you are already familiar with the basic properties of complex numbers.)
We make the following definition
eiθ^ = cos θ + i sin θ. (1)
This formula is called Euler’s Formula. In order to justify this use of the exponential notation appearing in (1), we will first verify the following form of the Law of Exponents: eiθ^1 +iθ^2 = eiθ^1 eiθ^2 (2)
To prove this we first expand the right-hand side of (1) by first multiplying out the product: eiθ^1 eiθ^2 = (cos θ 1 + i sin θ 1 )(cos θ 2 + i sin θ 2 ). Next we apply to this the trigonometric identities:
cos θ 1 cos θ 2 − sin θ 1 sin θ 2 = cos(θ 1 + θ 2 ) sin θ 1 cos θ 2 + cos θ 1 sin θ 2 = sin(θ 1 + θ 2 ).
When all this is done the result is
eiθ^1 eiθ^2 = cos(θ 1 + θ 2 ) + i sin(θ 1 + θ 2 ).
The right hand side of the last equation is exactly what we would get if we wrote out (1) with θ replaced by θ 1 + θ 2. We have therefore proved (2).
To justify the use of e = 2. 718.... , the base of the natural logarithm, in (1), we will differentiate (1) with respect to θ: We should get ieiθ^. Treating i like any other constant, we find (^) dθd eiθ^ = (^) dθd
cos θ + i sin θ
= − sin θ + i cos θ. But − sin θ + i cos θ = i(cos θ + i sin θ) = ieiθ. Thus, as expected,
d dθ
eiθ^ = i eiθ^ (3)
If one does not define eiθ^ by (1), then one must find some other mean to define ez^ and then to derive (1) directly as a consequence. Often the definition of ez^ is made using power series with complex numbers z but this requrires a considerable amount of preliminary work with power series. For a very brief discussion of this approach, see page 154 in the text.
Some examples: eiπ/^2 = i, eπi^ = − 1 , and e^2 πi^ = +1. Recall that the relation between the rectangular coordinates (x, y) and the polar coordinates (r, θ) of a point is
x = r cos θ, y = r sin θ
r =
x^2 + y^2 , θ = arctan
y x
where arctan (also called tan−^1 ) is one of the “branches” of the inverse tangent function. (The quadrant which holds the point (x, y) determines the correct branch of tan−^1 .) If z = 0 then r = 0 and θ can be anything. Making use of Euler’s formula, we can express polar representation in the following manner:
z = x + iy = r(cos θ + i sin θ) = reiθ, (5)
where r = |z| =
x^2 + y^2 and θ is given by (4). The angle θ is also called an argument of z and we write θ = arg(z).
As noted, there is an ambiguity in (4) about the inverse tangent formula for θ which can (and must) be resolved by looking at the signs of x and y in order to determine in which quadrant eiθ^ lies. For example, if x = 0, then the formula for θ in (4) makes no sense; but x = 0 simply means that z = 0 + iy lies on the imaginary axis so θ must be π/2 or 3π/2 depending on whether y is positive or negative. Again, if z = −4 + 4i, then r =
42 + 4^2 = 4
2 and θ = 3π/4. Therefore −4 + 4i = 4
2 e^3 πi/^4. Note also that, due to the periodicity of sin θ and cos θ, if z = reiθ, then we also have z = rei(θ+2kπ), k = 0, ± 1 , ± 2 ,.... Thus, in our last example, −4 + 4i = 4
2 e^11 πi/^4 = 4
2 e−^5 πi/^4 etc.
√ Here is another example:^ the complex number 2 + 8i^ may also be written as 68 eiθ, where θ = tan−^1 (4) ≈ 1 .33 rad. See Figure 1.
2 + 8i =
68 eiθ
θ = tan−^1 4 ≈ 1 .33 radians
Figure 1.
The conditions for equality of two complex numbers using polar coordinates are not quite as simple as they are for rectangular coordinates. If z 1 = r 1 eiθ^1 and z 2 = r 2 eiθ^2 , then
z 1 = z 2 if and only if r 1 = r 2 and θ 1 = θ 2 + 2kπ, k = 0, ± 1 , ± 2 ,...
Despite this the polar representation is very useful when it comes to multplication and division:
if z 1 = r 1 eiθ^1 and z 2 = r 2 eiθ^2 , then z 1 z 2 = r 1 r 2 ei(θ^1 +θ^2 ); (6)
z 1 z 2
= z 1 z 2 − 1 =
r 1 r 2
ei(θ^1 −θ^2 )^ (z 2 6 = 0). (7)
From the fact that
eiθ
)n = einθ^ we obtain De Moivre’s formula:
(cos θ + i sin θ)n^ = cos nθ + i sin nθ.
Expanding on the left and equating real and imaginary parts, leads to trigonometric identities which can be used to express cos nθ and sin nθ as a sum of terms of the form (cos θ)j^ (sin θ)k. For example with n = 2 one gets:
(cos θ + i sin θ)^2 = cos^2 θ − sin^2 θ + i 2 sin θ cos θ = cos 2θ + i sin 2θ.
Hence cos 2θ = cos^2 θ−sin^2 θ and sin 2θ = 2 sin θ cos θ. For n = 3, let us set C = cos θ and S = sin θ. Then (C + iS)^3 = C^3 + 3iC^2 S − 3 CS^2 − iS^3 , so
cos 3θ = Re {(C + iS)^3 } = C^3 − 3 CS^2 = 4 cos^3 θ − 3 cos θ.
(because S^2 = 1 − C^2 )). One can derive a similar identity for sin 3θ.
1 The exponential of any complex number.
The definition of ex+iy^ is given by the formula
ex+iy^ = exeiy^ (8)
Each term on the right-hand side of (8) already has a well defined meaning. It is left as an exercise to show that
d dt
e(a+bi)t^ = (a + bi)e(a+bi)t^ (9)
for any complex constant a + bi.
Exercises
- Let z 1 = 3i and z 2 = 2 − 2 i.
(a) Plot the points z 1 + z 2 , z 1 − z 2 , and z 2. (b) Compute |z 1 + z 2 | and |z 1 − z 2 |. (c) Express z 1 and z 2 in polar form.
- Let z 1 = 6eiπ/^3 and z 2 = 2e−iπ/^6. Plot z 1 z 2 , and z 1 /z 2.
- (a) Find and plot all complex numbers which satisfy z^3 = −8.
(b) Find all complex numbers z = reiθ, which satisfy z^2 =
2 eiπ/^4.
- Verify (9). Note that e(a+bi)t^ = eat^ · eibt. Use the product differentiation rule on this. You can differentiate eibt^ by means of (3) and the chain rule. You will still have some algebra to do to get the form on the right of (9).
- Find an identity for sin 3θ using n = 3 in De Moivre’s formula. Write your identity in a way that involves only sin θ and sin^3 θ if possible.
- This problem explains the first real use of complex numbers. A cubic equation can be tranformed into the form:
x^3 = 3px + 2q,
where p and q are constants by replacing x with ax + b and multiplying the cubic by a constant. The graph of the right side is a straight line which must cross the graph of x^3 and therefore there must be a (real) solution to the cubic. Cordano found a formula:
x =
q +
q^2 − p^3
q −
q^2 − p^3
Try finding the solution to x^3 = 6x + 6 using this formula (p = 2 and q = 3).
Here is where complex numbers arise: To solve x^3 = 15x + 4, p = 5 and q = 2, so we obtain: x = (2 + 11i)^1 /^3 + (2 − 11 i)^1 /^3.
Even though this looks like a complex number, it actually is a real number: the second term is the complex conjugate of the first term. Check that (2+i)^3 = 2+11i, and thus the solution is x = 4.
