Download COMPLEX NUMBERS

CS131
Part I, Number Systems
CS131 Mathematics for Computer Scientists II
Note 4
COMPLEX NUMBERS
There is no real number x satisfying the equation x 2 = −1. A solution can
be obtained by introducing a new number i which is assumed to satisfy
i 2 = −1. A complex number is then one of the form a + ib where a and b
are real numbers. Complex numbers are compared for equality, added and
multiplied using the following rules. If a, b, c, d ∈ R then
a + ib = c + id ⇔ a = c and b = d ,
(a + ib) + (c + id ) = (a + c) + i (b + d ),
(a + ib)(c + id ) = (ac − bd ) + i (bc + ad )
A complex number a + ib can be represented by an ordered pair (a, b) of
real numbers. The number a ∈ R is called the real part and b ∈ R is called
the imaginary part. The set of all complex numbers will be denoted by C.
We then associate i with the pair (0, 1) and write a pair of the form (a, 0)
as just a, giving
i 2 = (0, 1)(0, 1) = (−1, 0) = −1
and
(a, b) = (a, 0) + (0, b) = a + (0, 1)(b, 0) = a + ib
Addition and multiplication of these pairs is defined as follows:
(a, b) + (c, d ) = (a + c, b + d ),
(a, b)(c, d ) = (ac − bd , ad + bc)
(for any a, b, c, d ∈ R).
Problem. Find the real and imaginary parts of (2 + 3i )(1 − i )i
Solution. We have
(2+3i )(1−i )i = (2+3i )(i −i 2 ) = (2+3i )(1+i ) = 2+3i +2i +3i 2 = −1+5i
so the real part is −1 and the imaginary part is 5
Since a complex number is a pair of real numbers it can be represented by
a point in the plane. In this context we talk of the complex plane or Argand
diagram.
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If a, b ∈ R then the complex conjugate of a + ib is defined to be a − ib. The
complex conjugate of z ∈ C will be denoted by z . In the Argand diagram
taking the complex conjugate reflects the number in the real axis.
Examples. 5 + 3i = 5 − 3i ,
−1 − 2i = −1 + 2i ,
7 = 7,
−i = i
Properties of Complex Conjugation. For any z , w ∈ C:
z + w = z + w,
zw = z w ,
z /w = z /w ,
z = z,
z ∈R⇔z =z
The real part of z is (z +z )/2 and the imaginary part of z is (z −z )/2i
Example. If a, b ∈ R then the real and imaginary parts of the
complex number 1/(a + ib) can be found by multiplying the numerator
and denominator by the complex conjugate of a + ib:
1
a − ib
a − ib
a
b
=
=
=
−
i
a + ib
(a + ib)(a − ib) (a 2 + b 2 )
a2 + b2
a2 + b2
If x , y ∈ R and x + iy 6= 0 then we can express x and y in polar coordinates,
x = r cos θ,
y = r sin θ
giving
x + iy = r (cos θ + i sin θ),
where r =
p
x 2 + y 2 , and θ satisfies tan θ =
y
x
6
'$
y • x + iy = r (cos θ + i sin θ)
θ
xr
&%
The angle θ is called the argument of x + iy, the name principal argument
being used forp
the unique value satisfying −π < θ ≤ π.
The number x 2 + y 2 is called the modulus of x + iy and denoted by
| x + iy | . Geometrically it represents the distance between x + iy and the
origin of the complex plane.
4–2
(1)
(2)
(3)
(4)
(5)
(6)
Properties of the modulus. For any z , w ∈ C:
| z |=|√
z|
| z |= z z
z z =| z |2
| zw |=| z || w |
| z + w |≤| z | + | w |
(the triangle inequality)
| z | − | w | ≤| z − w |
The first four are easily verified by writing z = a + ib and w = c + id . To
prove the triangle inequality, we have
| z + w |2 = (z + w )(z̄ + w̄ ) = z z̄ + w z̄ + z w̄ + w w̄ =| z |2 + | w |2 +w z̄ + z w̄ ,
(| z | + | w |)2 =| z |2 + | w |2 +2 | z || w | .
Write w z̄ = u + iv then w̄ z = u − iv so w z̄ + w̄ z = 2u. Now
p
√
2 | z || w |= 2 | z || w̄ |= 2 | z w̄ |= 2 u 2 + v 2 ≥ 2 u 2 = 2 | u |≥ 2u = w z̄ +z w̄ .
Hence | z + w |2 ≤ (| z | + | w |)2 so | z + w |≤| z | + | w |.
The name ‘triangle inequality’ arises from the representation of complex
numbers as points in the plane as illustrated in the following diagram.
6
•z +
|z + w #
| #| w |
#*
#
|z |
#
-
6
•w *•z
-
w
If we multiply two complex numbers of modulus 1 expressed in polar form
we find
(cos θ + i sin θ)(cos φ + i sin φ)
= (cos θ cos φ − sin θ sin φ) + i (sin θ cos φ + cos θ sin φ)
= cos(θ + φ) + i sin(θ + φ)
Taking θ = φ gives (cos θ + i sin θ)2 = cos 2θ + i sin 2θ. This is a special case
of the following result.
De Moivre’s Theorem. For any integer n,
(cos θ + i sin θ)n = cos nθ + i sin nθ
4–3
De Moivre’s theorem can be proved by induction for n ≥ 0. It then follows
that
1
(cos θ + i sin θ)−n =
(cos θ + i sin θ)n
1
=
cos nθ + i sin nθ
cos nθ − i sin nθ
=
cos2 nθ + sin2 nθ
= cos nθ − i sin nθ
= cos(−nθ) + i sin(−nθ)
so the result also holds for negative values of n.
Problem. Find all complex numbers z with z 3 = 1.
Solution. Write z = r (cos θ + i sin θ). Since | z 3 |=| z |3 = 1 we have
r =| z |= 1. Now
z3 = 1 ⇒
⇒
⇒
⇒
⇒
(cos θ + i sin θ)3 = 1
cos 3θ + i sin 3θ = 1 = 1 + i 0
cos 3θ = 1, sin 3θ = 0 (equating real and imaginary parts)
3θ = 0, 2π, 4π, . . .
θ = 0, 2π/3, 4π/3, . . .
Taking θ = 0 gives z = 1. Taking θ = 2π/3 gives
2π
π
π
2π
+ i sin
= cos(π − ) + i sin(π − )
z = cos
3
3
3
3
so
√
π
π
1
3
z = − cos + i sin = − + i
.
3
3
2
2
The other value of z can be found by taking θ = 4π/3 or by noting that
since z 3 = 1 we also have z 3 = 1. Hence the other solution is
√
1
3
z =− −i
.
2
2
Fundamental Theorem of Algebra. Proved by Carl Gauss in 1799,
this states that every polynomial equation of degree n with complex coefficients ai , i = 0, . . . , n has exactly n solutions which, in general, are complex
numbers.. He showed that the polynomial
an z n + an−1 z n−1 + · · · + a2 z 2 + a1 z + a0
4–4
has n roots or zeros. λ1 , λ2 , . . . , λn (some of which may be identical) and
can therefore be written as
an (z − λ1 )(z − λ2 ) . . . (z − λn )
Hence
an z n + an−1 z n−1 + · · · + a2 z 2 + a1 z + a0 = 0 ⇔ z ∈ {λ1 , λ2 , . . . , λn }.
To solve a quadratic equation az 2 + bz + c = 0 we first find the square roots
r1 and r2 (which in general will be complex numbers) of the discriminant
−b + r2
−b + r1
and
.
b 2 − 4ac. The solutions of the equation are then
2a
2a
Other notations.
Some books (usually associated with electrical
engineering) use j instead of i for the imaginary unit to avoid confusion
with the common symbol for electrical current.
The complex conjugate of z is sometimes denoted by z ∗ instead of z .
ABSTRACT
Content
Complex
numbers,
definition,
properties
of
complex
numbers,
polar representation, properties of the complex modulus, De Moivre’s theorem, Fundamental Theorem of
Algebra.
This Note introduces
√ the idea of a complex number, a quantity consisting of a real (or integer) number
and a multiple of −1.
Complex numbers have become an essential part of pure and applied mathematics.
It is unfortunate that such numbers are called ’imaginary’ since they are no more or less real than any
other type of number. It is best to regard all numbers as abstractions and not to insist on arbitrary
connections with the real world.
This Note describes the algebraic properties of complex numbers and their representation in the complex
plane. We also look at the modulus function when applied to complex numbers and De Moivre’s theorem.
We use the latter to solve simple polynomials with complex roots.
History
Complex, or ’fictitious’ numbers, were first introduced in 16th century Italy as a consequence of a competition in algebra among leading Italian mathematicians.
Baron Augustin Louis Cauchy, [1789-1857] was a French mathematician who employed rigorous methods
of analysis. His prolific output included work on complex functions, determinants, and probability, and
on the convergence of infinite series. In calculus, he refined the concepts of the limit and the definite integral. Cauchy has the credit for 16 fundamental concepts and theorems in mathematics and mathematical
physics, more than any other mathematician. His work provided a basis for the calculus.
He provided the first comprehensive theory of complex numbers, a theory which contributed to the development of mathematical physics and, in particular, aeronautics.
In 1799 Gauss proved the Fundamental Theorem of Algebra: every polynomial of degree n with complex
coefficients has exactly n complex roots. Attempts had been made to prove this theorem for more than a
century.
Gauss helped develop vector analysis, a topic closely related to complex numbers and then attempted to
investigate yet another
√ type of number - hypercomplex numbers such as a + bi + cj + dk in which i , j , k
are ’variations’ on −1.
Hypercomplex numbers are rarely used now.
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