Download F COMPLEX NUMBERS

F COMPLEX NUMBERS
In this appendix, we review the basic properties of complex numbers. A complex number
is a number z of the form
z = a + bi
where a, b are real numbers and i represents a number whose square is −1, that is,
i 2 = −1
√
We write i = −1. The real numbers a and b are called the real part and imaginary
part of z, respectively. We use the notation
y (imaginary axis)
3i
−2 + 2i
Re(z) = a,
z = 3 + 2i
2i
i
−4 −3 −2 −1
x (real axis)
1
−i
2
3
4
−2i
5
−2 − i
2
−3i
FIGURE 1 The complex plane.
y
3i
z = 3 + 2i
2
2i
2
|z |
i
3
=
1
+2
2
3
FIGURE 2 The absolute value |z| is the
distance from z to the origin.
3i
i
−3 −2 −1
−2 − i
z = a + bi
2i
x (real axis)
−i
−2i
1
2
3
z = a − bi
Figure 3 shows the complex number z = −2 + i and its conjugate z = −2 − i. Observe
that z is the reflection of z across the real axis.
Algebraic operations
y (imaginary axis)
−2 + i
√
√
For example, z = 3 − 2i is a complex number with Re(z) = 3 and Im(z) = − 2.
A complex number is real if its imaginary part is zero. A complex number of the form
z = bi is said to be purely imaginary.
The set of all complex numbers is denoted by a boldface C. We represent C geometrically as the set of points in the Cartesian plane. The complex number z = a + bi
corresponds to the point (a, b) as in Figures 1 and 3. We call this plane the complex plane
(or, less commonly, an Argand diagram).
The absolute value |z| (also called the modulus) is defined as the distance from (a, b)
to the origin:
|z| = a 2 + b2
√
√
For example, |3 + 2i| = 32 + 22 = 13.
The complex conjugate (or simply “conjugate") of z = a + bi is the number z
defined by
x
4
Im(z) = b
4
z = a − bi
FIGURE 3 The conjugate z is the
reflection of z across the real axis.
We manipulate complex numbers using the usual rules of algebra, with one difference:
we replace i 2 by −1 whenever it occurs. The commutative, associative, and distributive
laws are all valid for complex numbers.
Addition and Subtraction These operations are performed by adding or subtracting the
real and imaginary parts separately:
(a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi) − (c + di) = (a − c) + (b − d)i
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APPENDIX F
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For example,
(3 + 2i) + (1 − 4i) = (3 + 1) + (2 − 4)i = 4 − 2i
√
√
π i − ( 2 + 3i) = − 2 + (π − 3)i
Multiplication To multiply two complex numbers, use the distributive and associative
laws together with the relations
i 2 = −1,
ib = bi
(b any real number)
For example:
(7i)(2 + 3i) = (7i)(2) + (7i)(3i) = 14i + 21i 2 = −21 + 14i
distributive law
Here is another example:
Here is the general formula for multiplying
complex numbers (it is not necessary to
memorize this formula):
(3 + 2i)(1 − 4i) = 3(1 − 4i) + 2i(1 − 4i)
= 3 − 12i + 2i − (2i)(4i)
(a + bi)(c + di) = (ac − bd) + (ad + bc)i
= 3 − 10i − 8i 2 = 3 − 10i + 8 = 11 − 10i
To describe division, we state the following properties of the conjugate. They can be
verified directly, by writing z = a + bi and w = c + di.
THEOREM 1 Properties of the Conjugate and Absolute Value
numbers z and w,
For any two complex
z+w =z+w
zw = z w
|zw| = |z| |w|
If z = a + bi, then
zz = |z|2 = a 2 + b2
1
Division To find the quotient z/w, multiply the numerator and denominator by the conjugate of the denominator:
z
zw
zw
=
=
w
ww
|w|2
E X A M P L E 1 Dividing Complex Numbers Write
15 + 5i
in the form a + bi.
3 − 4i
Solution Multiply numerator and denominator by the conjugate 3 + 4i and use (1):
15 + 5i
15 + 5i
3 + 4i
(15 + 5i)(3 + 4i)
=
=
3 − 4i
3 − 4i
3 + 4i
32 + 4 2
=
25 + 75i
(45 − 20) + (60 + 15)i
=
= 1 + 3i
25
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APPENDIX F
COMPLEX NUMBERS
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Every non-zero complex number z has a reciprocal, namely
z
z
zz
since z
=
=1
z−1 =
zz
zz
zz
E X A M P L E 2 Finding the Reciprocal Write the reciprocal of z = 5 + i in the form
a + bi.
Solution
5
1
5−i
5−i
1
1
=
=
=
= 2
− i
z
5+i
(5 + i)(5 − i)
26 26
5 + 12
Roots of Polynomials
Caution: the relation
√
√ √
a b = ab
is only valid if one or both a or b is positive. If
a < 0 and b < 0, this relation fails to hold (it is
off by a minus sign):
√
√
√
√
√
−2 · −3 = ( 2i)( 3i) = − 6
√
(−2)(−3) = 6
Mathematicians were first confronted with complex numbers when they began to inves√
tigate roots of polynomials. Recall that if a > 0, then a√has two real square √
roots ± a.
2
Thus, the roots of the equation x − a = 0 are x = ± a. By convention, a denotes
the positive square root. The √
negative number −a also has two square roots, namely, the
purely imaginary numbers ± a i:
√ 2
√
± ai = (± a)2 i 2 = a(−1) = −a
√
Said another way, the equation x 2 + a = 0 has two imaginary roots ± a i. For example,
the roots of x 2 + 5 = 0 are
√
√
x = ± −5 = ± 5i
√
√
By convention, when a > 0, −a denotes the square root ai.
More generally, the roots of a quadratic equation
ax 2 + bx + c = 0
are described by the famous quadratic formula
√
√
−b ± D
−b ± b2 − 4ac
=
x=
2a
2a
where D = b2 − 4ac is the discriminant of the equation. The roots are real if D ≥ 0 and
complex if D < 0. For example, the equation
x 2 + 3x + 5 = 0
has complex roots:
x=
−3 ±
32 − 4(1)(5)
3
=− ±
2
2
√
√
−11
11
3
=− ±
i
2
2
2
The quadratic formula was known in antiquity. Equivalent forms of the formula appear in Babylonian clay tablets dating to 1800 BC. However, mathematicians did not
immediately accept the notion of a complex number. They focused on the real roots and
ignored the complex roots. It was not until the 1500’s that similar formulas for cubic (degree three) and quartic (degree four) polynomials were discovered. These formulas were
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APPENDIX F
COMPLEX NUMBERS
not only much more complicated than the quadratic formula, they had a very surprising
feature which is illustrated by the cubic equation x 3 − 3x − 1 = 0. According to cubic
formula, one root of this equation is
1/3 −1/3
2
2
x=
+
≈ 1.8793
√
√
1 + −3
1 + −3
This root√is a real number, but it is expressed by a formula that involves the complex
number −3. So we find, surprisingly, that even if we wanted to restrict ourselves to real
roots of equations, we would still need complex numbers to write down those roots.
Since there was no escape, mathematicians came to accept that complex numbers are
legitimate mathematical entities that can be manipulated much like real numbers. They
figure prominently in the mathematics of the 18th century and had been fully integrated
into mathematics by the early 1800’s. However, the terms “real" and “imaginary" are a
reminder of the time when complex numbers were regarded with suspicion.
The formulas discovered in the 16th century showed that quadratic polynomials have
two roots, cubic polynomials have three roots, and quartic polynomials have four roots.
The suspicion grew that a polynomial of degree n has n complex roots. This was eventually
proved by Gauss in 1799 (at the age of 22). The precise form of this result is called the
Fundamental Theorem of Algebra. Since some roots of a polynomial may be multiple
roots (double roots, triple roots, etc), we cannot claim that a polynomial of degree n has
n distinct roots. Rather, it has n roots where each root is counted with its “multiplicity".
THEOREM 2 Fundamental Theorem of Algebra Every polynomial f (x) degree n
with complex coefficients has n complex roots. More precisely, if
f (x) = an x n + · · · + a1 x + a0
(an = 0)
then there exist n complex numbers α1 , ..., αn (unique up to order) such that
f (x) = an (x − α1 )(x − α2 ) · · · (x − αn )
The numbers αi are the roots of f (x). The multiplicity of a root α is defined as the
number of times α occurs in the list α1 , α2 , . . . , αn .
Polar Representation
y
z = a + bi
b = r sin
A complex number z = a + bi corresponds to the point (a, b) in the complex plane. We
may describe z in terms of the polar coordinates (r, θ ) of (a, b) (Figure 4). The radial
coordinate is the distance from the origin, and thus
r = |z|
r = |z|
a = r cos
x
FIGURE 4 Polar coordinates of a complex
number
The angular coordinate θ , called the argument of z, is denoted arg(z). As we see in
Figure 4, z = a + bi has the polar representation
z = |z|(cos θ + i sin θ)
where θ = arg(z).
The argument θ = arg(z) satisfies
b
(if a = 0)
a
However, arg (z) is not a uniquely defined angle. Since we are free to increase θ by an
integer multiple of 2π , the angles θ + 2π k for k an integer all serve as arguments for z.
tan θ =
APPENDIX F
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COMPLEX NUMBERS
We must keep in mind, therefore, that arg(z) really refers not to a single angle but rather
to a whole set of angles that differ by multiples of 2π . A common convention is to choose
θ in the interval (−π, π ]. Note also:
•
•
If z = a is real, then arg(z) = 0 if a > 0 and arg(z) = π if a < 0.
If z = bi is purely imaginary, then arg(z) = π/2 if b > 0 and arg(z) = −π/2 if
b < 0.
E X A M P L E 3 Find the polar representation of z = 4 + 5i.
Proof The radial coordinate is r = |z| =
√
√
42 + 52 = 29. The angular coordinate is
θ = tan−1
The polar representation is
z = 4 + 5i =
5
≈ 0.896
4
√
29(cos 0.896 + i sin 0.896)
The next theorem describes several properties of the argument. Especially important
is Eq. (2), expressing the behavior under multiplication: the argument of a product is the
sum of the arguments.
THEOREM 3 Properties of the Argument The argument satisfies the following properties:
arg(zw) = arg(z) + arg(w)
2
z
arg( ) = arg(z) − arg(w)
w
3
arg(z−1 ) = − arg(z)
arg(z) = 0
(if z is a real number)
4
5
Proof Let θ1 = arg(z) and θ2 = arg(w). We compute zw using the polar representation.
Note that the addition formulas for cosine and sine are used in the last two lines:
zw = |z|(cos θ1 + i sin θ1 ) |w|(cos θ2 + i sin θ2 )
= |z| |w|(cos θ1 + i sin θ1 )(cos θ2 + i sin θ2 )
= |z| |w| (cos θ1 cos θ2 − sin θ1 sin θ2 ) + i(cos θ1 sin θ2 + cos θ2 sin θ1 )
= |z| |w| cos(θ1 + θ2 ) + sin(θ1 + θ2 )
Therefore, arg(zw) = θ1 + θ2 which is equal to arg(z) + arg(w) as required. Next, observe
that since zz−1 = 1, we have
arg(zz−1 ) = arg(z) + arg(z−1 ) = arg(1) = 0
This implies Eq. (4). Finally, we may apply (2) and (4) to prove (3):
z
arg( ) = arg(zw−1 ) = arg(z) + arg(w −1 ) = arg(z) − arg(w)
w
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APPENDIX F
COMPLEX NUMBERS
The polar representation gives us a geometric interpretation of multiplication of complex numbers (Figure 5). If z1 and z2 , have arguments θ1 and θ2 , then the product z1 z2 is
the complex number of argument θ1 + θ2 that lies on the circle of radius |z1 ||z2 |.
y
z2
2
z1 z 2
1+
z1
2
1
| z1 z 2 |
FIGURE 5 Geometric picture of
x
multiplication of complex numbers.
The set of complex numbers of absolute value one make up the unit circle in the
complex plane (Figure 6(A)). Indeed, if |z| = 1, then z has radial component 1. Complex
numbers of absolute value one have polar representation
z = cos θ + i sin θ
To multiply two complex numbers of absolute value one, we need only add their arguments
(Figure 6(B)).
y
y
z = e i = cos + i sin
|z| = 1
z1 z 2
z2
1+
2
z1
2
1
FIGURE 6 Complex numbers of absolute
value one lie on the unit circle.
1
x
(A)
1
x
(B)
Euler’s Formula and the Complex Exponential
The formula
arg(zw) = arg(z) + arg(w)
has the same form as the logarithm law
ln(zw) = ln z + ln w
This is the first hint that complex numbers and
exponentials are related.
Euler discovered a deep and important connection between complex numbers and the
exponential function. As we know, ex is defined for every real number x. Euler realized
that ex is also meaningful if x is a complex number. For x = iθ , Euler’s formula reads
eiθ = cos θ + i sin θ
6
The complex number on the right-hand side has absolute value one. So according to Euler’s
formula, we obtain all complex numbers of absolute value one as values of ex for x = iθ .
Furthermore, if z = eiθ , then arg(z) = θ .
APPENDIX F
COMPLEX NUMBERS
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We may take Euler’s formula as a definition of the previously undefined quantity eiθ .
If so, we must justify this definition by showing that imaginary exponentials obey the law
of exponents:
eiθ1 eiθ2 = ei(θ1 +θ2 )
7
This follows directly from (2). Indeed, if z1 = eiθ1 and z2 = eiθ2 , then
|z1 z2 | = |z1 | |z2 | = 1
and
arg(z1 z2 ) = arg(z1 ) + arg(z2 ) = θ1 + θ2
We see that z1 z2 is the complex number of absolute value one with argument θ1 + θ2 .
Thus z1 z2 = ei(θ1 +θ2 ) as required.
A more direct and perhaps more satisfactory approach is to define ez for any complex
number z by the power series
ez =
∞ n
z
n=0
n!
=1+z+
z2
z3
+
+ ···
2!
3!
This expansion is valid if z is real. With some care, we can verify that it converges if z is
complex and satisfies
ez+w = ez ew
Setting z = iθ and separating eiθ into real and imaginary parts, we obtain
eiθ = 1 + (iθ ) +
(iθ )2
(iθ )3
(iθ )4
(iθ )5
(iθ )6
+
+
+
+
···
2!
3!
4!
5!
6!
iθ 3
θ4
θ5
θ6
θ2
−i
+
+i −
···
2!
3!
4!
5!
6!
θ4
θ6
θ3
θ5
θ2
+
−
+ ··· + i θ −
+
− ···
= 1−
2!
4!
6!
3!
5!
= 1 + iθ −
= cos θ + i sin θ
In the last step we use the Taylor series of cos θ and sin θ . For z = a + bi, we have
ez = ea+bi = ea eib = ea (cos b + i sin b)
As a first application of Euler’s formula, we show that the sine and cosine functions
can be expressed in terms of imaginary exponentials.
E X A M P L E 4 Verify the formulas
cos θ =
eiθ + e−iθ
,
2
sin θ =
eiθ − e−iθ
2i
Proof Since cos(−θ) = cos θ and sin(−θ) = − sin θ, we find that Euler’s formula
applied with −θ yields:
e−iθ = cos(−θ) + i sin(−θ) = cos θ − i sin θ
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APPENDIX F
COMPLEX NUMBERS
Therefore
eiθ + e−iθ = (cos θ + i sin θ) + (cos θ − i sin θ) = 2 cos θ
The first formula follows. The second formula follows similarly.
Eq. (7) has the following generalization to a product with n factors
eiθ1 eiθ2 · · · eiθn = ei(θ1 +θ2 +···+θn )
Aformal proof involves induction. This relation is true for n = 2. If it is valid for n = k − 1,
then it also valid for n = k by the following computation:
eiθ1 eiθ2 · · · eiθn = eiθ1 eiθ2 · · · eiθn−1 eiθn = ei(θ1 +θ2 +θn−1 ) · eiθn
= ei(θ1 +θ2 +θn−1 +θn )
Taking θ1 = θ2 = · · · = θn , we obtain DeMoivre’s Formula:
(eiθ )n = einθ
8
This formula is often written in the form
(cos θ + i sin θ)n = cos nθ + i sin nθ
9
We may use Euler’s Formula to describe the roots of a complex number. Recall that if
x is real, then (ex )1/n = ex/n . But notice that ex/n is a positive nth root of ex . There may
exist other nth roots. For example, if n is even, then −ex/n is also an nth root. When we
allow complex roots, we find that every non-zero complex number has n distinct complex
nth roots. To prove this, recall that if θ = arg(z), then for every integer k,
z = |z|ei(θ+2πk)
Therefore,
z1/n = |z|1/n e
i
θ +2π k
n
θ + 2π k
θ + 2π k + i sin
= |z|1/n cos
n
n
We obtain a distinct root for each of the values k = 0, 1, ..., n − 1. Graphically, these
roots are spaced equally along the circle of radius |z|1/n (see Figure 7).
z
y
5
5
+
+
2π
5
4π
5
5
x
Radius = | z | 1/5
FIGURE 7 The five roots z1/5 of z lie on
the circle of radius |z|1/5 with angular
spacing 2π/5
5
+
6π
5
5
+
8π
5
APPENDIX F
COMPLEX NUMBERS
A9
E X A M P L E 5 Find the cube roots of 7.
Solution Since z = −7 is real, θ = arg(7) = 0, and the cube roots of 7 are the three
complex numbers
2π k
2π k 71/3 cos
+ i sin
3
3
k = 0, 1, 2
Explicitly, the three cube roots of 7 are:
k = 0: 71/3
(the positive cube root)
k = 1: 71/3 (cos 2π
3 + i sin
2π
3 )
= 71/3 (− 12 +
k = 2: 71/3 (cos 4π
3 + i sin
4π
3 )
= 71/3 (− 12 −
√
3
2 i)
√
3
2 i)
Euler’s Formula and Calculus
In the chapter on second order differential equations, we use Euler’s formula to find
solutions of constant coefficient equations. In this subsection, we provide the background
for this application.
Given a complex number λ = a + bi, we define the complex exponential function:
f (t) = eλt = e(a+bi)t = eta (cos tb + i sin tb)
We shall take the domain of f (t) to be the set of all real numbers R (in a more advanced
treatment, we would consider the domain of all complex numbers). The values of f (t)
are complex numbers, so the range of f (t) is contained in C.
If λ = a is a real number, then the following formula holds by the Chain Rule:
d λt
e = λeλt
dt
This formula remains valid for all values of λ. Before we can verify this fact, we must
define what is meant by the derivative f (t). Write f (t) as a sum of real and imaginary
parts:
ta
f (t) = eta cos
sin tb = f1 (t) + if2 (t)
tb +i e f1 (t)
f2 (t)
The derivative is then defined by differentiating the real and imaginary parts separately:
f (t) = f1 (t) + if2 (t)
Equivalently, we may define the derivative as a limit of difference quotients:
f (t) = lim
h→0
f (t + h) − f (t)
h
This limit is a limit of complex values. By definition, this means that the following difference tends to zero in absolute value as h → 0:
f (t + h) − f (t) f (t) −
→0
h
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APPENDIX F
COMPLEX NUMBERS
THEOREM 4
For all complex numbers λ,
d λt
e = λeλt
dt
Proof If λ = a + bi, then
λeλt = (a + bi)(eta cos tb + ieta sin tb)
= aeta cos tb − beta sin tb + i beta sin tb + aeta sin tb
=
d ta
d ta
te cos tb + i
e sin tb
dt
dt
= f1 (t) + f2 (t)
=
d λt
e
dt
In our study of second order differential equations, we use the operator notation
d
D=
for the derivative. In this notation, Theorem 4 states:
dt
Deλt = λeλt
Applying Theorem 4 twice, we obtain
D 2 eλt = λ2 eλt
More generally, if we apply the operator
P (D) = aD 2 + bD + c
to the complex exponential function y = eλt , we obtain
P (D)eλt = (aλ2 + bλ + c)eλt = P (λ)eλt
Exercises
1. Find the conjugate z and absolute value |z| of the complex numbers
(a) −4 + 2i
(b) 4 + 2i
(c) −9i
(d) 1 − i
9.
11.
1
2i
10.
4
1 + 2i
2+i
10 − i
12.
1+i
1−i
2. Plot the complex numbers and their conjugates in the complex
plane: 2 + i, 3 − i, −4 + i.
In Exercises 13 - 20, write the polar expression reiθ in the form a + bi.
In Exercises 3 - 12, write the complex number in the form a + bi.
13. e 3 i
14. e
π
15. e2+ 2 i
16. e4+5i
π
17. e−2+ 4 i
18. eln 9−πi
19. (3e3i )(2e6i )
20. (e1+3i )(e2− 2 i )
3. (3 + 2i)(4 − 5i)
1
4. ( i)(−7i)
3
5. (7)(6i − 2)
1 4
6. (1 + i)(1 + i)( − i)
2 3
7.
1
5 + 2i
8.
1
−3 + i
π
5π
3 i
1
APPENDIX F
In Exercises 21 - 24, write the complex number in polar form reiθ .
21. 5 − 5i
22. 3 + 2i
23. 23 + 43 i
24. i
In Exercises 25 - 28, calculate the argument of the complex number.
25. −i
26. 5 − 5i
27. 3 − 2i
28. 4e6+ 3 i
π
29. What are the arguments of the four fourth roots of 1.
COMPLEX NUMBERS
A11
35. Apply De Moivre’s theorem with n = 3 to prove the identities
(a) cos 3θ = cos3 θ − 3 cos θ sin2 θ
(b) sin 3θ = 3 cos2 θ sin θ − sin3 θ
36. If a > 0, then ln(−a) does not exist as a real number. Show that
negative real numbers have complex logarithms by proving ln(−a) =
ln a + πi.
37. Let z = a + bi and w = c + di be complex numbers. Verify that
z+w =z+w
zw = zw
31. Find the three cube roots of 5.
√
√ √
38. Give an example where the equality a b = ab does not hold.
Under what conditions on real numbers a and b does it hold?
32. Find the three cube roots of i and 5i.
39. Use Euler’s formula to verify the identities
30. What are the arguments of the five fifth roots of 1.
33. Find the four fourth roots of 1.
34. Find the six sixth roots of 9.
cosh x = cos(ix),
sinh x = −i sin(ix)
40. Calculate f (0) and f (2) where f (t) = e(2+3i)t .