Download Complex Numbers: Basic Results The set of complex numbers (C) is

Complex Numbers: Basic Results
The set of complex numbers (C) is the set of all order pairs of real numbers (R) with a rule + for addition and a rule · for
multiplication. Algebraically, the triple (C, +, ·) is a field. This is a consequence of the definitions of + and · and because R
is a field.
1. Notation and Terminology: A complex number z is traditionally expressed as z = x + iy and not usually written
as an ordered pair of real numbers (x, y). When y = 0 we write z = x and call z real. If x = 0 we write z = iy and z
is often called imaginary. The symbol i, known as the imaginary
unit, is defined to be the principle square root of −1:
√
so i is a solution of the equation z 2 + 1 = 0 and we define −1 := i.
Equality of complex numbers:
(a) Two complex numbers z = x + iy and w = u + iv are equal if and only if their real and imaginary parts match.
That is, z = w if and only if x = u and y = v.
Addition and multiplication of complex numbers are defined in terms of addition and multiplication of real numbers:
(b) If z = x + iy and w = u + iv then their sum z + w := (x + y) + i(u + v).
(c) If z = x + iy and w = u + iv then their product z · w := (xu − yv) + i(xv + yu).
You should check that z + w = w + z, zw = wz and, for complex numbers a and b, z(a + b) = za + zb. And also
that products of three (or more) complex numbers, like abc, are unambiguous, i.e., that (ab)c = a(bc). Once these
properties are checked the definition for zw is not necessary to memorize as it can be obtained by expanding the product
(x + iy)(u + iv), using the relation i2 = −1 and grouping terms. Like real numbers we usually just write zw for the
product z · w.
A real number x can be regarded as a complex number through the identification x = x + 0i. So 0 = 0 + 0i and
1 = 1 + 0i. The complex numbers 0 and 1 are, in fact, the additive and multiplicative identities of C. That is, 0 + z = z
and 1z = z for each z ∈ C.
Additive inverses and quotients:
(d) If z = x + iy then its additive inverse is −z := (−x) + i(−y) =: −x − iy, so that z + −z = 0. Note that −1z = −z.
(e) The quotient of a complex number a with a nonzero complex number b is a complex number z such that bz = a.
In this case we write z = a/b. If a = 1 then z = 1/b and is called the reciprocal (multiplicative inverse) of b. The
reciprocal of b is often denoted by b−1 .
Exercise 1: Suppose z = x + iy, a = a1 + ia2 , b = b1 + ib2 6= 0 and bz = a. Equate real and imaginary parts to get
a1 b1 + a2 b2
a2 b1 − a1 b2
two equations involving x and y. Solve for x and y to show that z =
+i
. What are the real
b21 + b22
b21 + b22
and imaginary parts of the reciprocal 1/b?
Exercise 2: Show that 0z = 0 for any z and if wz = 0 then either w = 0 or z = 0.
2. The Complex Plane: The one-to-one correspondence between complex numbers and ordered pairs of real numbers
allows us to regard points in the plane as complex numbers and complex numbers as points in the plane: z = x + iy
is identified with the point having coordinates (x, y) and visa-versa. The horizontal x-axis is often called the real axis,
Re z, and the vertical y-axis the imaginary axis, Im z.
Figure 1: The Complex Plane
Im z
yi
• z = x + iy ∼ (x, y)
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i•
|z|
θ
•
1
x
Re z
The modulus (or length
por absolute value or magnitude) |z| of a complex number z is the Euclidean distance between
z and 0. Hence, |z| = x2 + y 2 . An argument of z, denoted by arg z, is an angle z makes with the positive real axis.
For example, arg(1 + i) = π/4 or 9π/4 or −7π/4 or . . . . Any two arguments for a complex number differ by an integer
multiple of 2π, so π/4 + 2kπi for any integer k is an argument for 1 + i. The interval −π < θ ≤ π of sometimes taken to
a the principle range of the argument function, but not always. If z = x + iy and x 6= 0 then tan(arg z) = y/x, which is
just the slope of the line through 0 and z. The formula θ = Tan−1 (y/x) returns an argument for z if x > 0. For x < 0
you’ll likely need adjust this by ±π becuase the arc-tangent function (at least on a calculator) usually returns a value
between −π/2 and π/2. For example, Tan−1 (y/x) + sign(y)π should work in this case — here, sign(t) is defined as 1
if t > 0, −1 if t < 0 and 0 if t = 0.
Exercise 3: Find a rule you could program on your (programmable) calculator that returns an argument θ of z such
that 0 ≤ θ < 2π. Assume that −π/2 < Tan−1 (t) < π/2 for any real number t (the principle range of the arc-tangent
function). Also assume that your calculator is smart enough to return Tan−1 (±∞) = ±π/2. Some software does this,
e.g., MatLab. Try out Tan−1 (1/0) and Tan−1 (−1/0) on your calculator and see what it does.
Addition and multiplication each have a geometric interpretation. The sum, w + z, of complex numbers w and z
corresponds to familiar vector addition in the plane. And multiplication of z by w will scale z by |w| and then rotate
the result through the angle arg z. This “scale-rotate” doesn’t seem obvious from the product definition (c) and so is
worth expanding on briefly (in and after the figure). However, once we define and get used to the complex exponential
function the “scale-rotate” of multiplication will seem rote.
Figure 2: w + z and wz
Im z
Im z
Multiplying z by w
Adding w to z
•w+z
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wz •
w
•
•z
θ
•z
Re z
•
|w|z
Re z
2
Scaling, then rotating (or visa-versa) is a linear
process in R . Multiplication of w and z is described in more detail
a
here using again the indentification a + ib ∼
, of C with R2 . If w = u + iv and z = x + iy then from definition (c)
b
ux − vy
u −v
x
u/|w| −v/|w|
x
cos θ − sin θ
x
wv = (ux − vy) + i(vx + uy) ∼
=
= |w|
= |w|
vx + uy
v u
y
v/|w| u/|w|
y
sin θ
cos θ
y
u/|w|
where θ is an argument of w since
∼ w/|w| is a unit vector in the direction of w.1
v/|w|
In short,
wz ∼
cos θ
sin θ
− sin θ
cos θ
x
|w|
∼ (scale z by |w|) then (rotate through θ = arg w)
y
3. Conjugation: The conjugate of a complex number z = x + iy is the complex number z̄ := x − iy. z̄ is the reflection
of z through the real axis:
Figure 3: Conjugation
Im z
• z = x + iy
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Re z
• z̄ = x − iy
Exercise 4: Check that
„
cos θ − sin θ
sin θ
cos θ
rotating it through an angle θ.
1 The
matrix
«
acts on R2 by rotating it back onto itself. That is,
2
„
cos θ
sin θ
− sin θ
cos θ
«„ «
„ «
x
x
is the vector obtained from
by
y
y
(a) z̄¯ = z.
(b) z z̄ = |z|2 .
(c) w + z = w̄ + z̄.
(d) wz = w̄z̄.
(e) z = z̄ if and only if z is real.
4. The Exponential Function: The power series 1 +
X zn
can be shown to converge absolutely for each z ∈ C and is a
n!
n≥1
natural way to define the complex number “ez ”. We’ll likely come back to this point of view later in the semester. But
we have yet to discuss converging sequences of complex numbers and so we’ll take another approach for the definition
ez , by using an initial value problem. We define eat to be the unique solution w of the initial value problem
ẇ(t) = aw(t), for t ∈ R
w(0) = 1,
(IVP)
even when a is a complex number. If a is a real number then we know from any elementary course in differential
equations that (IVP) defines the exponental function w(t) = eat , t ∈ R.2 But what if a is complex? In this case the
differential equation in (IVP) forces the values of w(t) = eat to also be complex, otherwise the left hand side of this
equation is real and the right hand side complex. Since the (IVP) is really all we have to work with it should determine
both the real and imaginary parts of w(t) (that is, of eat ). Let a = a1 + ia2 and w = u + iv then ẇ = u̇ + iv̇ and the
differential equation ẇ = aw splits into the coupled real system
u̇ = a1 u − a2 v
v̇ = a2 u + a1 v
(S)
Since w(0) = 1 then u(0) = 1 and v(0) = 0. This system together with the initial conditions on u and v forms an initial
value problem that also can be shown to have a unique solution (cf., footnote below). This shows that (IVP) is well
posed even if a is a complex number.
Special Case a = i: In this case w(t) = eit . But a = i also means a1 = 0, a2 = 1 and simplifies the coupling in system
(S):
u̇ = −v
v̇ = u.
Evidently then, ü = −v̇ = −u, and u(0) = 1 and u̇(0) = −v(0) = 0. The equation ü + u = 0 models an ideal
oscillating spring-mass system without friction or external forcing. The intial conditions u(0) = 1 and u̇(0) = 0 imply
that u(t) = cos t is the (unique) solution. Because v(t) = −u̇(t) = sin t then eit = w(t) = u(t) + iv(t) = cos t + i sin t.
This fundamental relation
eit = cos t + i sin t,
t∈R
(E)
it
is likely something you’ve seen before and
√ shows√the complex number e to be a point on the unit circle with argument
t. For example i = eπ/2i , −1 = eπi , −1/ 2 − i/ 2 = e5π/4i , and 1 = e2πi .
The cosine and sine functions both have period 2π. It then follows from (E) that eiθ also has period 2π: ei(θ+2π) = eiθ
for any θ ∈ R. This can also be seen by examining the next figure. More generally, ei(θ+2kπ) = eiθ for any θ ∈ R and
any integer k.
Figure 4: The Unit Circle
Im z
i
it
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t
1
Re z
Any complex number z has a polar form z = reiθ where r = |z| and θ is an argument of z. See figure 4.
2 Reference
any introductory text on differential equations. In fact, the initial value problem can be used to (numerically) compute values of et .
3
Figure 5: Polar Form of a Complex Number
Im z
iθ
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i.....
..... z = re ; r = |z|
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e •
θ
1
Re z
It’s left still to define ez , that is, ex+iy . This would be easy if the complex exponential function had the property
ea+b = ea eb
(A)
for complex numbers a and b (the real exponential function has this property). If such were the case then ez = ex+iy =
ex eiy = ex (cos y + i sin y) and ez would have Re (ez ) = ex cos y and Im (ez ) = ex sin y as real and imaginary parts.
Property (A) does in fact hold and is a consequence of (IVP). For suppose ẇa = awa , ẇb = awb and wa (0) = wb (0) = 1.
d
d
(wa wb ) = ẇa wb + wa ẇb = awa wb + bwa wb and the product wa wb satifies
(wa wb ) = (a + b)wa wb together
Then
dt
dt
with intial condition (wa wb )(0) = wa (0)wb (0) = 1. Uniqueness of solutions to (IVP) implies (wa wb )(t) = e(a+b)t for
any t ∈ R. Thus, e(a+b)t = (wa wb )(t) = wa (t)wb (t) = eat ebt and setting t = 1 is property (A).
Some further properties of the exponential function:
(a) e0 = 1.
(b) ez+2πi = ez for each complex number z (the exponential function has period 2πi).
(c) ez = 0 has no solution (the exponential function never vanishes).
(d) If w 6= 0 then ez = w has infinitely many solutions (each nonzero complex number has “logarithms”).
5. Complex Roots: If w ∈ C, n ∈ N, and z n = w then z is called an n-root of w. An n-th root
√ z of w is denoted
√ by
z = w1/n . For example, both i and −i are square roots of −1 since (±i)2 = −1 and 1, −1/2 + i 3/2, and −1/2 − i 3/2
are each a cube root of 1 (you should check this).
If z = reiθ and w = ReiΘ then z n = w means rn einθ = ReiΘ . Match modulii and arguments: rn = R and nθ = Θ, to
get z = R1/n eiΘ/n where R1/n is the usual (principle) real n-th root of the non-negative real number R.
However, we could have written w = Rei(Θ+2kπ) for any integer k. Let’s see where this approach leads: z n = w implies
Θ+2πk
rn einθ = Rei(Θ+2kπ) which implies z = zk := R1/n e n , k ∈ Z. It appears that there are infinitely many n-th roots:
z = zk for each integer k. Because the exponential function is periodic there are exactly n roots (you should convince
yourself of this):
z0 = R1/n eiΘ/n
z1 = R1/n eiΘ/n+i2π/n
..
.
zk = R1/n eiΘ/n+i2kπ/n
..
.
zn−1 = R1/n eiΘ/n+i2(n−1)π/n
and they are symmetrically distributed around the circle centered at 0 with radius R1/n .
4
Figure 6: z0 , z1 , and z2 : the 3 Cube Roots of 1 + i
Im z
√
6
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•1 + i
2 exp (iπ/12 + i2π/3) = z1 .•.......................
• z0 =
√
6
√
6
2 exp (iπ/12)
Re z
•
2 exp (iπ/12 + i4π/3) = z2
(Special Case: w = 1) z n = 1 has n solutions: z0 = 1, z1 = ei2π/n , . . . , zk = ei2kπ/n , . . . , zn−1 = ei2(n−1)π/n . The
numbers z0 , z1 , . . . , zn−1 are known as the n, nth roots of unity. Notice that z1k = zk for k = 0, 1, . . . , n − 1, that is, all
of the roots of unity can be generated by powers of the special root z1 . Such roots of unity are called primitive and
often given the special designator Ωn , i.e., Ωn = ei2π/n .
Figure 7: The Seven, 7th Roots of Unity
Im z
z •
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6
z3 •
• z = exp 2πi/7(=: Ω7 )
z =1
• 0
Re z
z4 •
z5 •
•z
Exercise 5: If zk is any one of the n, nth roots of unity and m is any integer then check that zkm is also an nth root
of unity.
Additional Exercises:
Exercise 6: Show that eiθ = e−iθ (So |z| = 1 if and only if z̄ = 1/z).
Exercise 7: Show that
i. |Re z| ≤ |z| and |Im z| ≤ |z|.
ii. |w + z| ≤ |w| + |z|.
iii. ||w| − |z|| ≤ |w − z|.
Exercise 8:
i. Find a number z so that ez = 1 + i. Hint: Write 1 + i in polar form.
ii. Given that w = u + iv 6= 0 then find a logarithm of w, i.e., a number z so that ez = w.
Exercise 9: If |w| ≤ 1 and |z| ≤ 1 then show that |w + z| ≤ |1 + w̄z|. When does equality hold?
Hint: Fix w in the unit disk. Use properties of conjugation (|a|2 = aā, ab = āb̄, a + ā = 2Re (a)) to show that |w + z|2 =
|w|2 + 2Re (w̄z) + |z|2 and |1 + w̄z|2 = 1 + 2Re (w̄z) + |w|2 |z|2 . All you need argue now is that
|w|2 + |z|2 ≤ 1 + |w|2 |z|2 ,
(A)
for then |w + z|2 ≤ |1 + w̄z|2 and taking square roots will finish the problem. To get inequality (A) set x = |z|2 and α = |w|2
and consider the graph of the simple function f (x) = (1 + αx) − (α + x) for 0 ≤ x ≤ 1. Note that equality holds in the
original inequality if and only if equality holds in (A) if and only if f (x) = 0.
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