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NOTE: the Lecture Notes general solution to the Diophantine equation
ax + by = k
looks a bit different to the one obtained in the lectures, but is equivalent, because in that case x0, y0 were solutions to
ax + by = gcd(a, b),
rather than to the problem we want to solve, namely
ax + by = k.
This explains the k/gcd(a, b) factor multiplying x0 and y0 in the general solution given in the book.
Use whichever approach you like, but be sure to be consistent!
COMPLEX NUMBERS
[This lecture: 3.1 plus starts of 3.2 and 3.3 in the Lecture Notes.
Basically anything not needing trigonometry!]
The first number system “invented” was the natural numbers 1, 2, 3, . . .,
or N.
The operations of addition and multiplication can be defined on N.
It made sense to extend this by allowing negative numbers and zero
as well: this gives the integers Z.
But then also fractions or rational numbers Q proved necessary, to
allow division.
The real numbers R are needed so that the “gaps” on the number line
√
are filled (e.g. 2 is irrational).
But mathematicians didn’t stop there!
To solve the equation x2 = −1 requires an extension of the real
numbers.
We know negative numbers can appear under the square root sign in
the quadratic formula. (“No real roots”.)
In fact the general solution to the cubic equation
ax3 + bx2 + cx + d = 0
can be given in a formula, but the formula involves square roots of
negative numbers even if all solutions are real!
So there is a need to have square roots of negative numbers too.
Allowing this proves very useful in: calculus, engineering, physics,
geometry, computer graphics, and many other areas.
A complex number is a “number” of the form
z = a + bi
where a, b are real and i is assumed to satisfy i2 = −1.
Then a is the “real part” of z and b is the “imaginary part”.
The set of all complex numbers is denoted by C.
We define addition, multiplication and subtraction by just using all the
usual laws of algebra and replacing i2 by −1 wherever it arises.
So we assume associativity and commutativity of both addition and
multiplication, as well as the distributive law.
Examples:
•
(2 + 3i) + (1 − 5i) = 2 + 1 + 3i − 5i
= 3 − 2i.
•
(2 + 3i)(1 − 5i) = 2(1 − 5i) + 3i(1 − 5i)
= 2 − 10i + 3i − 15i2
= 2 − 7i − 15(−1)
= 17 − 7i.
•
−(2 − 6i) = (−1)(2 − 6i)
= (−1)2 + (−1)(−6i)
= −2 + 6i.
In this way, we can deduce the general formulas for the sum or product of two complex numbers.
For addition:
(a + bi) + (c + di) = (a + c) + (b + d)i.
For multiplication:
(a + bi)(c + di) = a(c + di) + bi(c + di)
= ac + adi + bci + bdi2
= (ac − bd) + (ad + bc)i.
And for negatives:
−(a + bi) = −a − bi.
And for complex numbers z1 and z2, we define
z1 − z2 := z1 + (−z2).
Actually, these are usually taken to be the definitions of the complex
number operations, and the usual laws of algebra can be shown to
hold.
Notice that a copy of the real numbers R sits inside C: all complex
numbers with imaginary part 0.
This is because for complex numbers with zero imaginary part, the
complex number operations are the usual ones for real numbers.
(Of course this should be true if C really is an extension of R.)
So we call any complex number of the form a + 0i real.
Another example calculation.
Give (2 − 3i)(3 + i) − (1 + 2i) in rectangular form. (This means
expressed as a + bi where a, b ∈ R.)
(2 − 3i)(3 + i) − (1 + 2i) = 2(3 + i) − 3i(3 + i) − 1 − 2i
= 6 + 2i − 9i − 3i2 − 1 − 2i
= 6 − 7i + 3 − 1 − 2i
= (6 + 3 − 1) + (−7i − 2i)
= 8 − 9i.
Representing z = a + bi as an ordered pair (a, b) allows us to draw
z in the Argand plane.
We can view z = a + bi as a vector, which is an arrow having its tail
at the origin and its head at the point (a, b).
There is a nice interpretation of complex number addition as the “sum
of vectors”.
(More on general vectors later.)
The conjugate of z = a + bi is defined to be z̄ = a − bi.
On the Argand plane, the conjugate of z is obtained by reflecting
about the x-axis.
Properties of conjugation (all are easy to prove):
1. z̄¯ = z
2. z1 + z2 = z¯1 + z¯2
3. z1z2 = z¯1z¯2
Notice that z is real if and only if z = z̄.
Multiplying z by its conjugate is useful.
If z = a + bi,
zz̄ = (a + bi)(a − bi)
= a2 + abi − abi − b2i2
= a2 + b2.
So by Pythagoras’ Theorem, the length of the vector associated with
√
z is |z| = zz̄.
(More on the geometry of vectors later.)
For z ∈ C, we define |z| :=
√
zz̄.
We call |z| the modulus of z.
If z = a + bi, then |z| =
q
a2 + b2 (as we just proved).
So |z| is a real number, and non-zero providing z 6= 0.
If z ∈ C is real (so its imaginary part is zero), |z| is the usual absolute
value of z.
Example:
|3 + 4i| =
q
(3 + 4i)(3 − 4i)
q
32 + 4 2
p
9 + 16
=
=
=
√
25
= 5.
Properties of the modulus:
1. | − z| = |z|
2. |z1 + z2| ≤ |z1| + |z2| (triangle inequality)
3. |z1z2| = |z1| · |z2|
The first of these is easy to prove.
The second is really a fact about lengths of vectors in R2 and is obvious from a picture. It can be proved algebraically though. (We won’t
need it much.)
The important third property follows from the properties of conjugation:
|z1z2|2 = (z1z2)(z1z2)
= z1z2z¯1z¯2
= (z1z¯1)(z2z¯2)
= |z1|2|z2|2.
Taking square roots of both sides gives us what we want.
How about inverses?
Let z = a + bi 6= 0.
Then zz̄ = a2 + b2 6= 0 also.
1 z̄.
We define z −1 := |z|
2
Key property: z × z −1 = 1.
Proof:
z × z −1 = z ×
1
z̄
2
|z|
1
=
z × z̄
|z|2
=
1
2
|z|
|z|2
= 1.
So we call z −1 the inverse of z.
Example: find the inverse of 4 − 3i in rectangular form.
1
1
(4
−
3i)
=
(4 + 3i)
2
2
2
|4 − 3i|
4 +3
1
(4 + 3i)
=
16 + 9
=
1
(4 + 3i)
25
4
3
=
+
i.
25
25
(Can check by multiplying by 4 − 3i: is the result 1?)
In fact, now that we know about inverses of non-zero complex numbers, we can define “complex division”.
For z1, z2 ∈ C with z2 6= 0, we now define
z1/z2 := z1z2−1.
We can also compute this by doing something similar to “rationalising
the denominator”:
multiply by z̄2/z̄2 = 1.
Example: compute (−1 + 7i)/(2 + i).
−1 + 7i
−1 + 7i 2 − i
= (
)(
)
2+i
2+i
2−i
=
(−1 + 7i)(2 − i)
(2 + i)(2 − i)
−1(2 − i) + 7i(2 − i)
22 + 1 2
−2 + i + 14i + 7
=
5
5 + 15i
=
5
=
= 1 + 3i.
In fact this is a way to compute the inverse of z = a + bi also.
For example,
1
2+i
1
= (
)(
)
2−i
2−i 2+i
=
2+i
22 + 1 2
=
2+i
5
2
1
=
+ i.
5
5
We can do more complicated things too.
So if u, v, w are given complex numbers, we can solve
uz + v = w
for the unknown z.
Subtracting v from both sides gives:
uz = w − v,
and then dividing both sides by u gives:
z = (w − v)/u.
So all just ordinary algebra, followed by evaluation using the complex
number rules.