Download 3 The Field of Complex Numbers

3
The Field of Complex Numbers
Recall the standard description of the complex numbers:
C = {a + ib : a, b ∈ R, i =
√
−1}.
Any complex number z = a + ib can be thought of as having two components, a real part, Re(z) = a, and
an imaginary part, Im(z) = b. A real number is then just a complex number with imaginary part equal to
zero. A complex number is called pure imaginary if its real part is zero.
We add and multiply numbers in C as follows (compare this with the previous description of C).
(a + ib) + (c + id) = (a + c) + i(c + d) and (a + ib)(c + id) = (ac − ad) + i(ac + bd).
Then C is an ID, with zero 0 = 0 + 0i, unity 1 = 1 + 0i and the negative of z = a + ib is −z = (−a) + i(−b) =
−a − ib. (We leave it as an exercise to show that the associative (G1,R1) and distributive laws hold (R2),
and that both operations are well-defined and commutative.) Suppose we’d like to solve the equation
(2 + 3i)z = 4 − i. If (2 + 3i)−1 exists in C, then we can solve for unique z ∈ C by
z = (2 + 3i)−1 (4 − i).
Note that (a + ib)(a − ib) = a2 + b2 is a positive real number for any non-zero complex number z = a + ib.
Consider the following:
1
2 − 3i
2 − 3i
2
−3
2
−3
1
=
=
+i
=
=
+i
.
2
2
2
2 + 3i
2 + 3i
2 − 3i
4+3
4+3
4+3
13
13
Then (2 + 3i)−1 ∈ C, since
2 −3
13 , 13
∈ R. So we can solve for z, and in fact
z = (2 + 3i)−1 (4 − i) = (
2
3
5
14
− i )(4 − i) =
−i .
13
13
13
13
We can compute the inverse of an arbitrary complex number z = a + ib using the same technique. In general,
if z 6= 0 then a2 + b2 is a positive real number and
z −1 = (a + ib)−1 =
a2
b
a
−i 2
,
2
+b
a + b2
is well-defined in C. It follows that C is a field.
There is a particular map on C called complex conjugation:
θ : C → C : z = a + ib 7→ z̄ = a − ib.
This is an example of a special kind of map called a field automorphism. Note that
z z̄ = a2 + b2 ∈ R+
and so
z(a2 + b2 )−1 z̄ = 1 = zz −1 .
In particular, the inverse of z in C is a multiple of its conjugate, z̄, i.e.
z −1 =
z̄
.
a2 + b 2
The quantity a2 + b2 is the square of the modulus of a + ib.
Definition
3.1. Let z = a + ib be a complex number. The modulus of z is denoted by |z| and is given by
√
|z| = a2 + b2 .
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In this notation, we may write z −1 = z̄/|z|2 for any non-zero z ∈ C We note the following facts relating
to z and z̄
Lemma 3.1. Let z, z1 , z2 ∈ C. Then
1. z z̄ = |z|2 ,
2. z1 + z2 = z¯1 + z¯2 ,
3. z1 z2 = z¯1 z¯2 ,
4. cz = cz for any c ∈ R.
3.1
The Geometry of Complex Numbers
We’ve remarked before that z = a + ib can be identified with
√ the vector
√ (a, b) ∈ R × R. Note that the
2
2
Euclidean distance between (a, b) and the origin is given by a + b = z z̄, so the modulus of z, denoted
by |z|, can be interpreted as its distance to the origin in the complex plane.
Two complex numbers have the same modulus d iff they both lie on the same circle of radius d about
the origin.
The argument of a complex number, denoted arg(z), is the angle that the vector representing z makes
with the positive real axis, measured in the anti-clockwise direction.
Definition 3.2. Let z = a + ib ∈ C, let r = |z|, and let θ = arg(z). The polar form of z is given by
z = rcos θ + irsin θ.
If z = a + ib with a, b > 0 and arg(z) = θ then 0 ≤ θ ≤ π/2 and z is positioned in the 1st quadrant of
the argand diagram. Its relationship with the numbers ±a ± ib is shown in the table below.
z
a + ib
−a + ib
−a − ib
a − ib
arg(z)
θ
π−θ
π+θ
2π − θ
interval
[0, π/2]
[π/2, π]
[π, 3π/2]
[3π/2, 2π]
quadrant
1st
2nd
3rd
4th
polar form
r(cos θ + isin θ)
r(−cos θ + isin θ)
r(−cos θ − isin θ)
r(cos θ − isin θ)
Lemma 3.2. Let z ∈ C and let c be a positive real number. Then
1. |cz| = c|z|,
2. arg(cz) =arg(z)
√
Example 3.1. Let z = 2i − 2 3. Then |z| = 4 and arg(z) = 2π/3. The polar form of z is given by
z = 4(cos(2π/3) + isin(2π/3)).
It conjugate is
z̄ = 4(cos(2π/3) − isin(2π/3)) = 4(cos(4π/3) + isin(4π/3))
We recall the following trigonometric identities.
cos(A + B) = cosAcosB − sinAsinB
sin(A + B) = sinAcosB + cosAsinB
These can be applied for fast multiplication in C.
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Let z = r(cos A + sin A) and let w = (cos B + sin B). Then
zw = rs(cos(A + B) + sin(A + B)).
In words, the modulus and argument of the product of a pair of complex numbers is the product of their
moduli and the sum of their arguments, respectively.
This leads to the following theorem.
Theorem 3.1. (De Moivre’s Theorem) Let z = r(cosθ + sinθ) and let n be a non-negative integer. Then
z n = rn (cosnθ + sinnθ).
Proof. We apply an inductive proof. It’s clear the statement is true for the case n = 1. Suppose it is true
for n = k. Then
z k+1 = zz k = r(cosθ + sinθ)rk (coskθ + sinkθ) = rk+1 (cos(k + 1)θ + sin(k + 1)θ).
√
√
Example 3.2. Let z = 2i−2 3 and let w = 1+i. Then |z| = 4, |w| = 2, arg(z) = 2π/3, and arg(w) = π/4.
The product zw can be computed as
√
zw = 4 2(cos(2π/3 + π/4) + isin(2π/3 + π/4))
√
= 4 2(cos(11π/12) + isin(11π/12))
√
√
= −(2 + 2 3) + i(2 3 − 2).
De Moivre’s theorem is very useful for finding nth powers of a complex number.
(zw)6 = 46 23 (cos(11π/2) + isin(11π/2))
= 215 ((cos(π/2) + isin(π/2)))
= i215 .
Remark: Note that cos(2π+θ) = cosθ and sin(2π+θ) = sinθ. Also cos(π+θ) = −cosθ and sin(π+θ) = −sinθ.
3.2
Complex Roots of Unity
If z has modulus 1 then so does any power of z. Note that if z = cosθ + isinθ then z n = cosnθ + isinnθ from
De Moivre’s theorem, so if θ = 2πk/n for some non-negative integer k, we get z n = 1.
√
Example 3.3. Let z = 1/2 + i 3/2. Then |z| = 1 and arg(z) = π/6. The polar form of z is z =
cos(π/6) + isin(π/6). Then z 12 = cos2π + isin2π = 1. We say that z is a 12th root of unity.
We list the distinct powers of z as follows.
√
z
= cos(π/6) + isin(π/6)
= 1/2
+ i 3/2
√
3/2 + i1/2
z 2 = cos(π/3) + isin(π/3)
=
z 3 = cos(π/2) + isin(π/2)
= i √
z 4 = cos(2π/3) + isin(2π/3)
= − 3/2 +√i1/2
z 5 = cos(5π/6) + isin(5π/6)
= −1/2 + i 3/2
z 6 = cosπ + isinπ
= −1
= z 6 = −z 12
√
7
z
= cos(7π/6) + isin(7π/6)
= −1/2
− i 3/2 = z 5 = −z
√
z 8 = cos(4π/3) + isin(4π/3)
= − 3/2 − i1/2 = z 4 = −z 2
9
z
= cos(3π/2) + isin(3π/2)
= −i
= z 3 = −z 3
√
10
3/2 −√i1/2
= z 2 = −z 4
z
= cos(5π/3) + isin(5π/3)
=
z 11 = cos(11π/6) + isin(11π/6) = 1/2 − i 3/2
= z̄
= −z 5
12
12
= −z 6
z
= cos(2π) + isin(2π)
= 1
= z
10
Note that z 7 z 9 = z 16 = z 12 z 4 = z 4 since z 12 = 1. In fact every product z s z t = z s+t gives another complex
12th root of unity for any s, t ∈ {1, ..., 12}, so complex multiplication is an operation on the the set of all
such 12th roots of unity. It is not hard to see that the set
C12 = {z, z 2 , z 3 , ..., z 11 , z 12 = 1}
forms a group wrt complex multiplication: the identity element 1 is contained in C12 , and ck has inverse
c12−k in C12 for each k (for example z 3 z 9 = z 12 = 1).
Definition 3.3. Let n be a positive integer. A complex number z is called an nth root of unity if z n = 1.
We construct these nth roots as follows. Let z = r(cos(θ) + isinθ) be a complex number. Suppose that z is
an nth root of unity. Then
z n = rn (cos(nθ) + isin(nθ)) = 1 = cos(2kπ) + isin(2kπ),
for any non-negative integer k. Set r = 1. Then
zk = cos(2kπ/n) + isin(2kπ/n)
is an nth root of unity for each non-negative integer k.
Observe that zk = zt iff k ≡ t modulo n (i.e. iff k − t is a multiple of n). To see this, note that
z t = z k ⇔ z t z n−k = 1 ⇔ z n z t−k = z t−k = 1 ⇔ t − k = mn
for some integer m. This means that
z0 = zn = z2n = · · · = ztn = · · · ,
z1 = zn+1 = z2n+1 = · · · = ztn+1 = · · · ,
..
.
zn−1 = z2n−1 = · · · = ztn−t+1 = · · · ,
so in particular there are exactly n distinct complex nth roots of unity.
The set of these nth roots is given by
Cn = {z1 = z, z2 = z 2 , z3 = z 3 , ..., zn−1 = z n−1 , zn = z n = 1}.
Lemma 3.3. The complex nth roots of unity form a group under complex multiplication.
Proof. If z, w are both complex nth roots of unity then
(zw)n = z n wn = 1 · 1 = 1,
so the product of any pair of complex nth roots of unity is another nth root, and hence complex multiplication
is an operation on the set of all such nth roots of unity. Multiplication is assoc in C, and hence on any
subset of C, so G1 is immediately satisfied. The identity 1 is trivially an nth root of unity for any positive
integer n, so G2 holds. Finally, z has inverse z n−1 for any complex nth root of unity z so G3 is satisfied.
Remark: Cn is a non-empty subset of C\{0}, and is also a group wrt complex multiplication. We thus say
that Cn is a subgroup of C\{0}. We’ll examine substructures of groups, rings and fields in the next section.
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