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Complex Numbers and
the Complex Exponential
Frank R. Kschischang
The Edward S. Rogers Sr. Department of
Electrical and Computer Engineering
University of Toronto
September 15, 2005; updated January 10, 2017
1 Numbers and Equations
Numbers have often been invented to solve equations.
For example, by introducing negative numbers the natural numbers N = { 0 , 1 , 2 ,.. .} can be extended to the integers Z = {... , − 2 , − 1 , 0 , 1 , 2 ,.. .} so that we may solve simple equations such as x+2 = 0. Likewise, by introducing integer quotients, the integers can to be extended to the rational numbers Q = {a/b : a, b ∈ Z, b 6 = 0} so that we may solve simple equations such as 2x = 1.
The rationals seem like a very nice set in which to do arithmetic. There is a well-defined ad- dition operation and a well-defined multiplication operation, and Q is closed with respect to these operations. Addition and multiplication are associative (i.e., for all x, y, z ∈ Q, x(yz) = (xy)z and likewise for addition) and commutative (i.e., for all x, y ∈ Q, x + y = y + x and likewise for multiplication). Every element x ∈ Q has an additive inverse −x (from which we may define a subtraction operation) and every nonzero element x ∈ Q, x 6 = 0, has a mul- tiplicative inverse 1/x (from which we may define a division operation). Furthermore, multi- plication and addition satisfy the distributive law, i.e., for all x, y, z ∈ Q, x(y + z) = xy + xz. Mathematically speaking, the rational numbers form a field. Who could ask for anything more?
The trouble is that certain simple equations such as
x^2 − 2 = 0 (1)
have no solutions in Q, i.e., no rational number x satisfies (1). However, following the progression from N to Z to Q, we might try to get around this problem by extending Q, i.e., by adjoining an element—let’s call it θ for now—that satisfies θ^2 − 2 = 0, or, equivalently, θ^2 = 2. We will demand that θ be combinable with ordinary rational numbers (and with itself) via addition and multiplication, while satisfying all of the formal arithmetic properties (such as closure with respect to addition and multiplication, associativity, commutativity, the distributive law, etc.) that we have grown to expect.
If we denote this extended set by Q[θ], then certainly Q[θ] must contain all numbers of the form a + bθ, where a, b ∈ Q. Numbers involving higher powers of θ do not arise, since any such higher power can be reduced to a multiple of a lower power, i.e., θ^2 = 2, θ^3 = 2θ, θ^4 = 4, etc. Indeed, the sum, difference, product and quotient of any two elements of this form is another element of this form (provided that we don’t attempt to divide by zero). To see this, observe that (a + bθ) ± (c + dθ) = (a ± c) + (b ± d)θ,
and since a ± c and b ± d are rational when a, b, c and d are rational, we have another number of the same form. Likewise
(a + bθ)(c + dθ) = ac + adθ + bcθ + bdθ^2 = (ac + 2bd) + (ad + bc)θ.
Again, since ac + 2bd and ad + bc are rational when a, b, c and d are rational, we have another number of the same form. Finally, to show that we can form quotients, it is enough to show that we can form reciprocals. Note that if a and b are not both zero then
1 a + bθ
a − bθ (a + bθ)(a − bθ)
=
a − bθ a^2 − b^2 θ^2
=
a a^2 − 2 b^2
b a^2 − 2 b^2
θ.
Now, a^2 − 2 b^2 6 = 0 (why?), so a/(a^2 − 2 b^2 ) and −b/(a^2 − 2 b^2 ) are rational when a, b, c and d are rational; thus once again we have another number of the same form. Indeed, we have shown that Q[θ] is a field, and unlike Q, in this field (1) does indeed have a solution.
Since every element of Q[θ] can be written as a + bθ, with a, b ∈ Q, we could associate such an element with the ordered pair (a, b). We might refer to the first component, a, as the “rational part” of a + bθ and the second component, b, as the “irrational part” of a + bθ. (Note that the “irrational part” is in fact itself a rational number!) The arithmetic in Q[θ] could be defined via operations that occur on such ordered pairs, e.g.,
(a, b) + (c, d) = (a + c, b + d) (a, b) × (c, d) = (ac + 2bd, ad + bc)
We usually denote individual complex numbers with a single symbol, such as z. Now, since every such element can be written in the form z = a + b j, with a, b ∈ R, we can associate such an element with the ordered pair (a, b) of real numbers. We refer to the first component, a, as the “real part” of z (written Re(z)) and the second component, b, as the “imaginary part” of z (written Im(z)). Thus for all z ∈ C, we have
z = Re(z) + Im(z) j.
Note that the “imaginary part” is in fact itself a real number! If Im(z) = 0, then z is referred to as a (purely) real number; likewise if Re(z) = 0, then z is referred to as a purely imaginary number. Thus, in particular, j itself is purely imaginary.
Two complex numbers z 1 and z 2 are equal if and only if their real and imaginary parts agree, i.e., z 1 = z 2 if and only if [Re(z 1 ) = Re(z 2 )] and [Im(z 1 ) = Im(z 2 )].
Thus every complex equality can always be regarded as a pair of real equalities.
The complex conjugate z∗^ of a complex number z = Re(z) + Im(z) j is the complex number
z∗^ = Re(z) − Im(z) j.
Note that z∗^ = z if and only if Im(z) = 0, i.e., if and only if z is real.
The complex conjugate obeys the following properties for all w, z ∈ C:
(w ± z)∗^ = w∗^ ± z∗; (wz)∗^ = (w∗)(z∗); (w/z)∗^ = (w∗)/(z∗) (z 6 = 0).
Furthermore, for all z ∈ C we have
z + z∗^ = 2 Re(z); z − z∗^ = 2 j Im(z); (z∗)∗^ = z.
It is possible to show (see the problems) that if f (x) = a 0 + a 1 x + a 2 x^2 + · · · + adxd^ is a polynomial with real coefficients (i.e., a 0 , a 1 ,... , ad ∈ R) and f (z) = 0 for some complex value z, then we must also have f (z∗) = 0. In other words, if z is a zero of a polynomial with real coefficients then so is its complex conjugate z∗.
Note that, unlike the real numbers, complex numbers are not in general ordered, i.e., it makes no sense to ask which is larger: 2 + 3 j or 3 + 2 j. However, we can always compare the magnitudes of two complex numbers. The magnitude (or absolute value or modulus) |z| of a complex number z is defined as
|z| =
zz∗^ =
Re^2 (z) + Im^2 (z).
x
y
unit circle
j
2 + 3 j
2 − 3 j
− 4 − 4 j
x
y
a + b j = r(cos θ + j sin θ)
r
a
b
θ
(a) (b)
Figure 1: (a) Several points in the complex plane. (b) The polar form of a complex number.
Clearly |z| is a non-negative real number, and |z| = 0 if and only if z = 0. Note that |z∗| = |z|, i.e., a complex number and its complex conjugate have the same magnitude.
It is very convenient to visualize C as a two-dimensional vector space over R, i.e., as a plane. Naturally, this plane is referred to as the complex plane. The number a + b j corresponds to the point with coordinates (a, b) in the complex plane, as shown in Fig. 1(a).
In Figure 1(a), the two axes have been labelled x and y. Since each point on the x-axis represents a real number, this axis is called the real axis. Similarly, the y-axis is called the imaginary axis. These two axes intersect at 0, which is the only complex number that is simultaneously purely real and purely imaginary.
The magnitude |z| of a complex number z has a geometric interpretation in the complex plane: |z| measures the (Euclidean) distance between the origin 0 and z. Or if z is regarded as a vector in the complex plane, then |z| is the length of this vector. Similarly, z∗^ represents the geometric reflection of z in the real axis.
3 Polar Form
The set of points at unit distance from the origin in the complex plane, corresponding to the complex numbers z with |z| = 1, form a circle of unit radius centered at the origin. This circle is called the unit circle in the complex plane. Every point on the unit circle can be
Now, if z 6 = 0, we have z(1/z) = 1; this means that if z has the polar form z = r(cos θ+j sin θ), then 1/z must have the polar form
1 z
r
[cos(−θ) + j sin(−θ)],
so that their product has the polar form 1 = 1(cos 0 + j sin 0). Thus taking reciprocals in polar form is just as convenient as complex multiplication. It follows from this that if z 2 6 = 0
arg(1/z 2 ) = − arg(z 2 ), |z 1 /z 2 | = |z 1 |/|z 2 |,
and arg(z 1 /z 2 ) = arg(z 1 ) − arg(z 2 ), provided z 1 6 = 0.
The property that the phase of a product is the sum of the phases is very reminiscent of the rule for multiplying exponentials, where the exponent of a product is the sum of the exponents, i.e., exey^ = ex+y. As we will see in the next section, where we consider the complex exponential function, this connection is not a coincidence.
4 The Complex Exponential
To define a complex exponential function ez^ , we would certainly wish to mimic some of the familiar properties of the real exponential function; e.g., ez^ should satisfy, for all z 1 , z 2 and z,
ez^1 ez^2 = ez^1 +z^2 ez^1 /ez^2 = ez^1 −z^2 d dz
ez^ = ez^.
(The last equation requires that we first define what we mean by complex differentiation, something that is beyond the scope of these notes. However, see, e.g., [1, 2].) Here we will introduce the complex exponential in a sneaky way: via its Maclaurin series. (Recall that the Maclaurin series of a function is the Taylor expansion of the function around zero.)
Recall that the real-valued functions cos(x), sin(x), and ex^ have Maclaurin series given, respectively, by
cos(x) = 1 −
x^2 2!
x^4 4!
x^6 6!
sin(x) = x −
x^3 3!
x^5 5!
x^7 7!
ex^ = 1 + x +
x^2 2!
x^3 3!
and that each of these series is convergent for every value of x ∈ R.
It would be natural indeed to define
ez^ = 1 + z +
z^2 2!
z^3 3!
∑^ ∞
i=
zi i!
and this is precisely what we will do. Although we will not prove this here, the Maclaurin series (4) is convergent for every value of z ∈ C. Clearly this complex-valued exponential agrees with the usual real-valued exponential at every point z on the real-axis in the complex plane.
To see that ez^1 ez^2 = ez^1 +z^2 , we multiply the corresponding Maclaurin series. This method of multiplying series is sometimes called the Cauchy product. We want to form
ez^1 ez^2 =
z^01 0!
z 11 1!
z^21 2!
z 13 3!
×
z 20 0!
z^12 1!
z 22 2!
z^32 3!
Using the distributive law and grouping terms having the same total exponent (sum of z 1 exponent and z 2 exponent), we get
ez^1 ez^2 =
z 10 z 20 0!0!
[
z 11 z 20 1!0!
z^01 z 21 0!1!
]
[
z 12 z^02 2!0!
z 11 z 21 1!1!
z^01 z 22 0!2!
]
[
z^31 z^02 3!0!
z 12 z^12 2!1!
z 11 z 22 1!2!
z^01 z 23 0!3!
]
This sum can be written as
ez^1 ez^2 =
∑^ ∞
i=
∑^ i
j=
zi 1 − jzj 2 (i − j)!j!
∑^ ∞
i=
i!
∑^ i
j=
i!z 1 i− jzj 2 (i − j)!j!
∑^ ∞
i=
i!
∑^ i
j=
i j
zi 1 − jzj 2
∑^ ∞
i=
(z 1 + z 2 )i i! = ez^1 +z^2 ,
where, in the second last equality, we have made use of the binomial expansion (which holds in any field).
Now, writing z = a + j b, with a ∈ R and b ∈ R, we find that
ez^ = ea+j^ b^ = eaej^ b.
Exercises
- Prove that no rational number x satisfies (1).
- Work out the rules of arithmetic for elements in Q[
3].
- Let f (x) be any polynomial in x with coefficients in C. It can be shown that every non-constant polynomial with coefficients in C, i.e., every function
f (x) = a 0 + a 1 x + a 2 x^2 + · · · + adxd
with a 0 , a 1 ,... , ad ∈ C and d > 0 has a zero in C, i.e., f (x) = 0 for some x ∈ C. This means that we cannot find new elements that satisfy simple polynomial relations to adjoin to C. Mathematically speaking, the complex field C is “algebraically closed.” Show that the fact that C is algebraically closed implies that every polynomial f (x) of degree d > 0 factors into a product of d degree-one polynomials with complex coefficients. (Hint: show that if f (x 0 ) = 0, then f (x) = (x − x 0 )g(x) where g(x) is a polynomial of degree d − 1. Then apply the principle of mathematical induction.)
- Let f (x) be a polynomial with real-valued coefficients. Show that if f (z) = 0 for some complex number z, then f (z∗) = 0 as well. (Hint: consider [f (z)]∗.)
- A complex-valued function f (z) is said to be periodic with period z 0 if f (z + z 0 ) = f (z) for all z ∈ C. Show that ez^ is periodic with period 2π j.
- Show that (cos θ+j sin θ)n^ = cos(nθ)+j sin(nθ), a result known as De Moivre’s formula.
- Use De Moivre’s formula to show that
cos(3θ) = cos^3 (θ) − 3 cos(θ) sin^2 θ
and sin(3θ) = 3 cos^2 (θ) sin(θ) − sin^3 (θ)
- Show that the complex exponential takes on every complex value except 0.
References
[1] E. B. Saff and A. D. Snider, Fundamentals of Complex Analysis for Mathematics, Science, and Engineering, 3rd Edition, Prentice-Hall, 2003.
[2] D. A. Wunsch, Complex Variables with Applications, 3rd Edition, Addison-Wesley, 2004.
Spoiler Alert
Don’t read further unless you want to see the solutions to the exercises.
deg g(x). Note that the remainder is either identically zero, or has a degree strictly smaller than that of g(x). Now, let f (x) be a polynomial with coefficients in C, and suppose f (x 0 ) = 0 for some x 0 ∈ C. Let g(x) = x − x 0. Note that deg g(x) = 1. By the division property described above, we can find q(x) and r(x) such that
f (x) = q(x)(x − x 0 ) + r(x),
with r(x) ≡ 0 or deg r(x) < 1. In either case r(x) = a 0 for some complex number a 0. However, evaluating f (x) at x = x 0 yields f (x 0 ) = 0 = q(x 0 )(x 0 − x 0 ) + r(x 0 ) = r(x 0 ), so a 0 = 0. This shows that f (x) = q(x)(x − x 0 ). If f (x) has degree d > 0, then q(x) must have degree d − 1 (since degrees add under polynomial multiplication). Now if f (x) is a polynomial over C of degree one, then the claimed property is obviously true, since f (x) = f (x) is a “factorization” of f (x) into a “product” of degree-one polynomials. Now assume, for d ≥ 1, that every polynomial of degree d over C factors as a product of d degree-one polynomials. Let f (x) be a polynomial of degree d + 1 over C. Since C is algebraically closed, f (x) has a zero, i.e., f (x 0 ) = 0 for some x 0 ∈ C. By our previous result, this implies that
f (x) = q(x)(x − x 0 ) (6)
for some polynomial q(x) of degree d. By our hypothesis, however, q(x) itself factors as a product of d degree-one polynomials. Substituting this factorization for q(x) in (6), we obtain a factorization of f (x) as a product of d + 1 degree-one polynomials. Since the claimed property holds for all degree-one polynomials, and we have just shown that whenever the property holds for all polynomials of degree d ≥ 1 it also holds for all polynomials of degree d + 1, it follows from the Principle of Mathematical Induction that the property holds for all d ≥ 1.
- Let f (x) = a 0 + a 1 x + a 2 x^2 + · · · + adxd^ be a polynomial with a 0 , a 1 ,... , ad ∈ R. Note that [f (x)]∗, the complex conjugate of f (x) can be written as
[f (x)]∗^ = (a 0 + a 1 x + a 2 x^2 + · · · + adxd)∗ = a∗ 0 + a∗ 1 x∗^ + a∗ 2 (x^2 )∗^ + · · · + a∗ d(xd)∗ = a 0 + a 1 x∗^ + a 2 (x∗)^2 + · · · + ad(x∗)d = f (x∗),
where, in the second last equality we have used the property that a∗ i = ai and (xi)∗^ = (x∗)i. Thus if f (z) = 0 for some z, then f (z∗) = [f (z)]∗^ = 0 also.
- For all z ∈ C we have ez+2π^ j^ = ez^ e^2 π^ j^ = ez^ , since e^2 π^ j^ = cos(2π) + j sin(2π) = 1.
- We have (cos θ + j sin θ)n^ = (ej^ θ)n^ = ej^ nθ^ = cos(nθ) + j sin(nθ).
- From De Moivre’s formula, we have
cos(3θ) + j sin(3θ) = (cos θ + j sin θ)^3 cos^3 (θ) + 3 cos^2 (θ) j sin(θ) + 3 cos(θ) j^2 sin^2 (θ) + j^3 sin^3 (θ) = cos^3 (θ) − 3 cos(θ) sin^2 (θ) + j[3 cos^2 (θ) sin(
Equating real and imaginary components yields the desired identities.
- Let z = rej^ θ^ with r > 0. Then z = eln^ rej^ θ^ = eln^ r+j^ θ, which shows that z is in the range of the complex exponential. On the other hand, if a + b j is a complex value with a, b ∈ R, then ea+b^ j^ = eaej^ b^ has magnitude ea, which is nonzero. Thus 0, which has magnitude 0, is not in the range of the complex exponential.
