Download MTH 4104 Introduction to Algebra 2 Complex numbers

MTH 4104
Introduction to Algebra
Notes 2
2
2.1
Spring 2017
Complex numbers
Basic definitions
The historical reason we need to consider number systems larger than the set of real
numbers R is because there are still equations we can’t solve, such as x2 = −1 (which
has no real solution) or x3 = 2 (which has only one, though for various reasons we
would like it to have three). It turns out that the first equation is the crucial one.
A complex number is an expression of the form a + bi, where a and b are real
numbers, and i is a mysterious symbol which will have the property that i2 = −1. The
rules for addition and multplication are
(a + bi) + (c + di) = (a + c) + (b + d)i,
(a + bi)(c + di) = (ac − bd) + (ad + bc)i.
You can work out the rule for subtraction. How do we divide? You can check that the
rule above gives
(a + bi)(a − bi) = a2 + b2 ,
which is a positive number unless a = b = 0. So, to divide by a + bi, we multiply by
b
a
− 2
i.
a2 + b2
a + b2
Thus, in the complex numbers, we can add, subtract, multiply, and divide (except
by zero). The laws of arithmetic which we are familiar with in the real context all
apply here too.
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Example
(2 − 3i)(4 − i)
2 − 3i
=
4+i
42 + 12
5 − 14i
=
,
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which can be verified by multiplying the result by 4 + i.
Complex numbers are not called complex because they are complicated: a modern advertising executive would certainly have come up with a different name! The
intended meaning of the name “complex” is “compound; made up of parts”. Each
complex number is built of two parts, each of which is simpler (being a real number).
If z = a + bi is a complex number (where a and b are real), we say that a and b
are the real part and imaginary part of z respectively. So the rules for addition and
subtraction can be put like this:
To add or subtract complex numbers, we add or subtract their real parts
and their imaginary parts.
The complex number a − bi is called the complex conjugate of z, and is written as z.
The rule for multiplication looks more complicated as we have written it out. There
is another representation of complex numbers which makes it look simpler. Let z =
a + bi. We define the modulus and argument of z by
p
a2 + b2 ,
|z|
=
arg(z) = θ where cos θ = a/|z| and sin θ = b/|z|.
In other words, if |z| = r and arg(z) = θ , then
z = r(cos θ + i sin θ ).
For example, let z = 1 + i. Then the modulus of z is
p
√
|z| = 12 + 12 = 2,
√
√
and the argument θ satisfies cos θ = 1/ 2 and sin θ = 1/ 2, so that θ = π/4.
Now the rules for multiplication and division are:
To multiply two complex numbers, multiply their moduli and add their
arguments. To divide two complex numbers, divide their moduli and subtract their arguments.
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2.2
The complex plane, or Argand diagram
The complex numbers can be represented geometrically, by points in the Euclidean
plane (which is usually referred to as the Argand diagram or the complex plane for this
purpose. The complex number z = a + bi is represented as the point with coordinates
(a, b). Then |z| is the length of the line from the origin to the point z, and arg(z) is the
angle between this line and the x-axis. See Figure 1.
r
z = a + bi
|z| = r b = r sin θ
θ
r
0
a = r cos θ
Figure 1: The Argand diagram
In terms of the complex plane, we can give a geometric description of addition and
multiplication of complex numbers. The addition rule is the same as you learned for
adding vectors in Geometry I, namely, the parallelogram rule (see Figure 2).
r z1 + z2
* r
z1
r
z2
*
r
0
Figure 2: Addition of complex numbers
Multiplication is a little bit more complicated. Let z be a complex number with
modulus r and argument θ , so that z = r(cos θ + i sin θ ). Then the way to multiply an
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arbitrary complex number by z is a combination of a stretch and a rotation: first we
expand the plane so that the distance of each point from the origin is multiplied by r;
then we rotate
√ the plane through an angle θ . See Figure 3, where we are multiplying
by √
1 + i = 2(cos(π/4) + i sin(π/4)); the dots represent the stretching out by a factor
of 2, and the circular arc represents the rotation by π/4.
(3 + 2i)(1 + i)
= 1 +.r 5i
.................
.............
............
...........
.........
.........
........
........
.......
.......
.......
......
......
.....
....
....
....
....
....
....
....
..
.
...
.
.
.
r. .
3 + 2i
r
0
Figure 3: Multiplication of complex numbers
Now let’s check the correctness of our rule for multiplying complex numbers.
Remember that the rule is: to multiply two complex numbers, we multiply the moduli
and add the arguments. To see that this is correct, suppose that z1 and z2 are two
complex numbers; let their moduli be r1 and r2 , and their arguments θ1 and θ2 . Then
z1 = r1 (cos θ1 + i sin θ1 ),
z2 = r2 (cos θ2 + i sin θ2 ).
Then
z1 z2 = r1 r2 (cos θ1 + i sin θ1 )(cos θ2 + i sin θ2 )
= r1 r2 ((cos θ1 cos θ2 − sin θ1 sin θ2 ) + (cos θ1 sin θ2 + sin θ1 cos θ2 )i)
= r1 r2 (cos(θ1 + θ2 ) + i sin(θ1 + θ2 )),
which is what we wanted to show.
From this we can prove De Moivre’s Theorem:
Theorem 2.1 For any natural number n, we have
(cos θ + i sin θ )n = cos nθ + i sin nθ .
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Proof The proof is by induction. Starting the induction is easy since (cos θ +i sin θ )0 =
1 and cos 0 + i sin 0 = 1.
For the inductive step, suppose that the result is true for n, that is,
(cos θ + i sin θ )n = cos nθ + i sin nθ .
Then
(cos θ + i sin θ )n+1 = (cos θ + i sin θ )n · (cos θ + i sin θ )
= (cos nθ + i sin nθ )(cos θ + i sin θ )
= cos (n + 1)θ + i sin (n + 1)θ ,
which is the result for n + 1. So the proof by induction is complete.
Note that, in the second line of the chain of equations, we have used the inductive hypothesis, and in the third line, we have used the rule for multiplying complex
numbers.
The argument is clear if we express it geometrically. To multiply by the complex
number (cos θ + i sin θ )n , we rotate n times through an angle θ , which is the same as
rotating through an angle nθ .
2.3
Polynomial equations over C
Given any reasonable number system1 and two numbers α, β in this system, we can
consider the linear equation
αz + β = 0,
to be solved for z. Provided α is non-zero, this equation has a unique solution, namely
β
z=− .
α
To see that this is true, we can solve the equation in the usual way, but taking care on
the way to note what operations we are performing, and to make sure that our number
system allows these operations, so that we’re not doing anything illegal. Very briefly:
αz + β
(αz + β ) + (−β )
αz
α −1 (αz)
z
=
=
=
=
=
0
−β
−β
α −1 (−β )
α −1 (−β ) = − αβ .
1 Those
⇒
⇒
⇒
⇒
who have read ahead can take “reasonable number system” to mean “field”. It is a worthwhile exercise to see which other field laws we are implicitly using: for instance, can you spot the
invocations of the associative laws?
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So, in our number system, we need to be able to add the negative of β to both sides of
the equation, and then we need to be able to divide the resulting equation by α, or put
another way, multiply both sides by the multiplicative inverse α −1 = α1 of α.
In C we can do both of these operations; therefore, all linear equations with α
nonzero have a solution in C.
What about quadratic equations? Let’s keep things more concrete this time, and consider the general quadratic equation
αz2 + β z + γ = 0
with complex coefficients α, β , γ ∈ C. Can we solve this equation inside the complex
numbers?
Of course “the answer” must be to use the quadratic formula
p
−β ± β 2 − 4αγ
z=
2α
which we know from School. But why does it work and what does it even mean?
There is only one way to be sure — and that’s to give a proof. The idea in the proof of
the correctness of the quadratic formula is that we can complete the square, as follows:
2
αz
+ β z + γ
z2 + β z + αγ
α 2
γ
β
z2 + αβ z + 4α
2 + α
2
β
z + 2α
= 0
⇒
= 0
⇒
=
=
β2
4α 2
β2
4α 2
⇒
− αγ =
β 2 −4αγ
4α 2
.
So far we have not done anything other than divide through by α (which is only legal
provided that α 6= 0 — if instead α = 0 then the argument is wrong), and add some
constants to both sides of the equation. Since the usual laws of arithmetic hold for
complex numbers, we are reasonably confident that everything so far is correct.
Now comes the difficult
step — we have to extract a square root. Provided we can do
p
2
this, and that by β − 4αγ we mean any complex number u satisfying the equation
u2 = β 2 − 4αγ, we can complete the proof as follows:
2
2 −4αγ
β
z + 2α
= β 4α
⇒
2
q
2 −4αγ
β
z + 2α
= ± β 4α
⇒
√ 2
2
−β ± β −4αγ
z =
.
2α
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So we see that the quadratic formula reduces solving quadratic equations over C to
the problem of extracting square roots inside C. This is best done by applying De
Moivre’s Theorem, but sometimes there are shortcuts.
Example Solve z2 + (1 − i)z − i = 0 for z ∈ C.
Soluton We apply the quadratic formula with α = 1, β = 1 − i and γ = −i. First
calculate the discriminant
β 2 − 4αγ = (1 − i)2 − 4(−i) = 1 − 2i + i2 + 4i = 1 + 2i + i2 = (1 + i)2
which is visibly a square. The equation u2 = (1+i)2 has exactly two solutions u = 1+i
and u = −(1 + i) in C (can you prove that this is true?) so
z=
−(1 − i) ± (1 + i) −1 + i ± (1 + i)
=
.
2
2
The plus sign gives the solution z = i and the minus sign gives the solution z = −1.
Here is the reason why the complex numbers are such a wonderful number system:
every polynomial equation with complex coefficients can be solved inside C! More
precisely:
Theorem 2.2 (Fundamental Theorem of Algebra) Let n ≥ 1, and let a0 , a1 , · · · , an−1
be complex numbers. The polynomial equation
zn + an−1 zn−1 + · · · + a1 z + a0 = 0
has at least one solution inside C.
Despite the name, the proof of this theorem is beyond the scope of this module.
You will see a proof in the module Complex Variables. However see Problem Sheet 2
for a proof of this Theorem for cubic equations, the ones with n = 3.
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