Download 2.4 Complex numbers (slides, 4to1)

Complex numbers
The complex number system is an extension of the real number system.
It unifies the mathematical number system and explains many
mathematical phenomena.
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√
We introduce a number i = −1 defined to satisfy the equation x2 = −1.
(Of course if i2 = −1 then x = −i also satisfies x2 = −1.)
Part 2, Polynomials
Lecture 2.4a, Complex Numbers
The complex numbers are defined as all numbers of the form a + bi
Write
Dr. Ken W. Smith
C := {a + bi : a, b ∈ R}.
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A complex number of the form
2013
z = a + bi
is said to have real part < = a and imaginary part = = b.
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Motivation for the complex numbers
Motivation for the complex numbers
Any “number” can be written as a complex number in the form a + bi.
The nicest version of the Fundamental Theorem of Algebra says that every
polynomial of degree n has exactly n zeroes.
The number 3i = 0 + 3i has real part 0 and is said to be “purely
imaginary”;
the number 5 = 5 + 0i has imaginary part 0 and is “real”.
But this is not quite true. Or is it?
Consider the functions f (x) = x2 − 1, g(x) = x2 and h(x) = x2 + 1. We
graph these functions below.
The real numbers are a subset of the complex numbers.
The conjugate of a complex number z = a + bi is created by changing
the sign on the imaginary part: z̄ = a − bi.
Thus the conjugate of 2 + i is 2 + i = 2 − i;
√
√
√
the conjugate of 3 − πi is 3 − πi = 3 + πi.
The conjugate of i is ī = −i and the conjugate of the real number 5 is
merely 5.
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Motivation for the complex numbers
Motivation for the complex numbers
It is obvious that the quadratic graphed in green
f (x) = x2 − 1 = (x − 1)(x + 1)
has two zeroes. Move the green parabola up one unit: g(x) = x2 . What
happened to our two zeroes? They merged into the single x-intercept at
the origin. We claim that g(x) = x2 still has two zeroes, if we are willing
to count multiplicities. This makes some sense because we can write
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g(x) = Elementary
x2 = (xFunctions
− 0)(x − 0)
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Motivation for the complex numbers
f (x) = x2 − 1 = (x − 1)(x + 1)
g(x) = x2 = (x − 0)(x − 0)
What if we move the parabola up one more step and graph h(x) = x2 + 1?
Now, suddenly, there are no solutions. The graph never touches the x-axis.
Suddenly we have lost our pair of solutions! Can this be fixed?
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Motivation for the complex numbers
Algebraically, h(x) = x2 + 1 does not have any real zeroes because that
requires
x2 + 1 = 0 =⇒ x2 = −1.
If we square any positive real number, the result is positive. So we cannot
get −1. But if we use imaginary numbers then the equation
x2
f (x) = (x − 1)(x + 1)
+1=0
2
still has
two
zeroes, i and −i. The
quadratic
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(SHSU)
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Functions x + 1 now factors as
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Motivation for the complex numbers
Motivation for the complex numbers
f (x) = (x − i)(x + i)
f (x) = (x − 0)(x + 0)
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Motivation for the complex numbers
Motivation for the complex numbers
In the late middle ages, mathematicians discovered that if one were willing
to allow for a new number, one whose square was −1, quite a lot of
mathematics got simpler!
Modern cell phone signals rely on sophisticated signal analysis; we would
not have cell phones without the mathematics of complex numbers.
More analysis of electrical wiring and electrical signaling uses complex
numbers.
(They particularly noticed that they could solve quadratic and cubic
equations!)
This “imaginary” number was therefore very useful.
Complex numbers appear throughout all of mathematics and greatly
simplify many mathematical problems!
Over time, the term “imaginary” has stuck, even though scientists and
engineers now use complex numbers all the time.
In the next presentation we will look at complex numbers in quadratic
equations.
It is now common agreement to write i as an entity that satisfies
(END)
i2 = −1.
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Complex numbers in Quadratic Equations
Complex numbers appear naturally in quadratic equations.
Suppose we wish to solve the quadratic equation
Elementary Functions
ax2 + bx + c = 0
Part 2, Polynomials
Lecture 2.4b, Complex Numbers in Quadratic Equations
By completing the square we can solve for x and find that
x=
Dr. Ken W. Smith
The expression b2 − 4ac under the radical sign is called the discriminant
of the quadratic equation and is often abbreviated by ∆.
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If ∆ = b2 − 4ac is positive then the square root of ∆ is a real number and
so the quadratic equation has two real solutions:
x=
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√
−b± b2 −4ac
2a
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Complex numbers in Quadratic Equations
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√
−b+ ∆
2a
and x =
√
−b− ∆
.
2a
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Complex numbers in Quadratic Equations
The quadratic equation
ax2 + bx + c = 0
The quadratic equation
has solutions
ax2
+ bx + c = 0
x=
√
If ∆ = b2 − 4ac is negative then ∆ is imaginary and so our solutions are
complex numbers which are not real.
√
√
To be explicit, if ∆ is negative then −∆ is positive and so ∆ = −∆ i.
has solutions
x=
√
−b± b2 −4ac
2a
If ∆ = b2 − 4ac is zero then there is only one solution since
√
x=
−b± ∆
2a
√
−b± b2 −4ac
2a
The solutions to the quadratic formula are then
√
b
= − b±2a 0 = − 2a
.
x=
This single solution occurs with multiplicity two.
√
−b+ −∆ i
2a
and x =
√
−b− −∆ i
2a
In this case, the plus/minus sign (±) in front of
will get two complex numbers as solutions.
√
∆ assures us that we
These two complex solutions come in conjugate pairs.
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Complex numbers in Quadratic Equations
Complex numbers in Quadratic Equations
Example. The solutions to the quadratic equation
x2
are
√
−1±
12 −4(1)(1)
3
=
√
−1± −3
2
+x+1=0
√ √
−1± 3 −1
2
=
Thus the two solutions to the equation
conjugate pairs
− 12
√
3
2 i
+
Since these are the two zeroes of
factor
= (x +
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−
x2
=
√
−1± 3 i
2
√
1± −3
2
√
±
3
2 i.
√
3
2 i))(x
+
1
2
1
2
+ x + 1 then we can
− (− 12 −
√
3
2 i
+
3
2 i))
1
2
= (x −
3
2 i)
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Complex numbers in Quadratic Equations
1
2
√
3
2 i.
±
√
1
2
and
√
3
2 i))(x
(x − ( 12 +
√
Elementary Functions
=
3
2 i.
−
then the polynomial x2 − x + 1 factors as
√
+
√
1± 3 i
2
=
Since the two solutions to the equation x2 − x + 1 = 0 are the complex
numbers
√
and − − 23 i.
the polynomial x2
√
3
2 i)(x
=
− 12
+ x + 1 = 0 are the complex
1
2
x2 + x + 1 = (x − (− 12 +
1
2
Some worked examples.
Solve the quadratic equation x2 − x + 1 = 0. Also, factor x2 − x + 1.
Solution. By the quadratic formula the solutions to x2 − x + 1 = 0 are
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−
− ( 12 −
√
3
2 i)(x
−
1
2
√
3
2 i))
√
3
2 i)
+
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Complex numbers in Quadratic Equations
Use the roots of 2x2 + 5x + 7 to factor 2x2 + 5x + 7.
Solution. Since the two solutions to the equation 2x2 + 5x + 7 = 0 are
Solve the quadratic equation
x = − 54 +
2x2 + 5x + 7 = 0
Solution. According to the quadratic formula,
√
√ √
√
−5± 52 −4(2)(7)
−5± −31
−5± 31 −1
x=
=
=
=
4
4
4
√
31
4 i
and x = − 54 −
√
−5± 31 i
4
= − 54 ±
(x − (− 54 +
√
31
4 i.
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√
31
4 i
and x = − 54 −
√
31
4 i.
2(x − (− 54 +
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√
31
4 i))(x
− (− 54 −
√
31
4 i))
must be a factor of 2x2 + 5x + 7. But if we check the leading coefficient
of the polynomial in the expression above, we see that we need to multiply
by 2 to complete the factorization. So 2x2 + 5x + 7 factors as
= 2(x +
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31
4 i
and since c is a zero of a polynomial if and only if x − c is a factor, then
Our two solutions are the conjugate pairs
x = − 54 +
√
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5
4
√
31
4 i))(x
− (− 54 −
√
−
31
4 i)(x
+
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5
4
√
31
4 i))
√
+
31
4 i)
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Complex numbers in Quadratic Equations
Elementary Functions
In the next presentation, we explore the algebra and geometry of complex
numbers.
Part 2, Polynomials
Lecture 2.4c, The Geometry and Algebra of Complex numbers
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Dr. Ken W. Smith
Sam Houston State University
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Geometric interpretation of complex numbers
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Complex numbers
Mathematicians began to recognize the value of complex numbers
sometime back in the Renaissance period (fifteenth and sixteenth
centuries) but it was not until there was a geometric interpretation of the
complex numbers that people began to feel comfortable with them.
We may view the complex numbers as lying in the Cartesian plane. Let
the traditional x-axis represent the real numbers and the traditional y-axis
represent the numbers of the form yi.
We equate a complex number x + yi with the point (x, y). (So the
imaginary numbers yi are “perpendicular” to the real numbers!)
(See the complex plane is drawn on the next page.)
The complex plane
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Smith (SHSU)
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Complex numbers
Complex numbers
The Argand Diagram
The process of changing a + bi into the point (a, b) can be traced to
Argand around 1800 and is sometimes called the “Argand diagram”. In
the Argand diagram, the complex number z = a + bi is equated with the
point (a, b) in the Cartesian plane. For example, 6 + 5i can be graphed as
the point (6, 5).
The ordinary Cartesian plane then becomes a plane of complex numbers.
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Thus i = 0 + 1i is equated with the point (0, 1) and the number 1 = 1 + 0i
is equated with the point
√ (1, 0). The point (2, 1) represents√the number
2 + i. The number ( 3 + i)/2 is equated with the point ( 3/2, 1/2).
In the complex plane the x-axis is called the “real” axis and the y-axis is
called the imaginary” axis.
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The algebra of complex numbers
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Complex numbers
We multiply complex numbers (a + bi)(c + di) just as we would the
polynomials (a + bx)(c + dx) (except that we remember that i2 = −1.)
The complex numbers have a natural addition, subtraction and
multiplication.
We add and subtract complex numbers just as we would polynomials,
keeping up with the real and imaginary parts. For example,
For example, since
(3 + 4x)(7 + 11x) = 21 + 61x + 44x2 .
(3 + 4i) + (7 + 11i) = 10 + 15i
then
and
(3 + 4i) − (7 + 11i) = −4 − 7i.
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(3 + 4i)(7 + 11i) = 21 + 61i + 44i2 = 21 + 61i − 44 = −23 + 61i.
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Division of complex numbers
Some worked examples.
We would like our complex numbers to be written in “Cartesian form”
a + bi so there is a little twist involved in doing division with complex
numbers.
Note that if z = a + bi then z z̄ = (a + bi)(a − bi) = a2 + b2 . So if we are
dividing by z, we may view z1 as
Write the complex fractions below into the “Cartesian” form z = a + bi
where a, b ∈ R.
1
2
z̄
a
1
b
=
= 2
− 2
i.
2
z z̄
a +b
a + b2
z
3
4
3+2i
7−3i .
5
3+i
2
1+i
1
i
Solution.
Computationally, this means that anytime we have a fraction involving a
complex number z in the denominator, we multiply numerator and
denominator by z̄ and simplify.
1
We want to be ready to use complex numbers when factoring polynomials
or solving polynomial equations. For example, let’s factor the polynomial
f (x) = x3 − 2x2 + 9x − 18.
Since this is a cubic polynomial and we don’t want to use the cubic
formula (there is a cubic formula but it is quite messy!) then we need to
find a zero.
We could try some numbers (see some techniques in the next lectures)
and discover by trial-and-error that f (2) = 0. Or we could graph this
polynomial on a graphing calculator and see that x = 2 is a zero.
Since f (2) = 0 (check this!) then x − 2 is a factor of x3 − 2x2 + 9x − 18.
Now divide x − 2 into f (x) = x3 − 2x2 + 9x − 18 by synthetic division:
1
2
1
−2
9
− 18
2
0
18
0
9
0
We seeSmith
that
f (x) factors as (x Elementary
− 2)(x2Functions
+ 9). Since x2 + 9 = 0 implies
that
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=
(3+2i)(7+3i)
(7−3i)(7+3i)
=
the imaginary part is
For example,
3 + 4i 7 + 11i
(3 + 4i)(7 + 11i)
−23 + 61i
−23
61
3 + 4i
=
·
=
=
=
+i
.
7 − 11i 7 + 11i
(7 − 11i)(7 + 11i)
170
170 170
7 − 11i
This process, multiplying the numerator and denominator of a fraction by
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the conjugate of the denominator, is called rationalizing the
denominator.
Using
complex numbers and the factor theorem
3+2i
7−3i
2
3
4
5
3+i
5
3−i
= ( 3+i
)( 3−i
)=
15+23i
58
23
58 .)
15−5i
10
=
2(1−i)
2
2
1−i
1+i = ( 1+i )( 1−i ) =
2
1 −i
−i
1
i = ( i )( −i ) = 1 = −i .
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=
3
2
15
58
+
23
58 i
. (So the real part is
15
58
and
− 12 i .
= 1−i .
(Or 0 − i .)
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Some worked examples.
Factor the polynomial f (x) = x3 − 8 and then find all solutions to x3 = 8.
Solution. Since f (x) = x3 − 8 is zero when x3 = 8 then we know that
one zero is x = 2. Since f (2) = 0 we know that x − 2 is a factor of f (x).
Dividing x3 − 8 by x − 2 gives us x3 − 8 = (x − 2)(x2 + 2x + 4). The
solutions to x3 − 8 = 0 are the solutions to (x − 2)(x2 + 2x + 4) = 0. By
the quadratic formula,
the solutions
to x2 + 2x + 4 = 0 are
√
√
1
x = 2 (−2 ± 2 3i) = −1 ± 3i.
This implies that (x2 + 2x + 4) factors as
√
√
√
√
x2 +2x+4 = (x−(−1+ 3 i))(x−(−1− 3 i) = (x+1− 3 i)(x+1+ 3 i).
So the factoring of f (x) is
x3 − 8 = (x − 2)(x + 1 −
√
√
3 i)(x + 1 + 3 i) .
The full set of solutions to the equation x3 = 8 is the set of solutions to
the equation x3 − 8 = 0. These are the zeroes of f (x):
√
√
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x = 2, x = −1
+ 3Functions
i, x = −1 − 3 i .
Factoring x6 − 1
Another example
Factor completely g(x) = x6 − 1.
Solution.The Fundamental Theorem of Algebra tells us the x6 − 1 has six
zeroes and therefore factors into six linear pieces.
One way to begin factoring g(x) = x6 − 1 is to view this polynomial as a
difference of squares:
g(x) = x6 − 1 = (x3 )2 − 12 = (x3 + 1)(x3 − 1).
To further factor x3 − 1, notice that x = 1 is surely a zero and after
synthetic division we see that
x3 − 1 = (x − 1)(x2 + x + 1).
In a similar way we should notice that x = −1 is a zero of x3 + 1 and so
g(x) = x6 − 1 = (x3 + 1)(x3 − 1) = (x + 1)(x2 − x + 1)(x − 1)(x2 + x + 1).
We are not done; we need to factor the two quadratics x2 − x + 1 and
x2 + x + 1. In an early example, we factored these using the quadratic
formula and found that
√
√
1
3
1
3
2
x + x + 1 = (x − −
i)(x − +
i)
2
2
2
2
and
√
√
1
3
1
3
i)(x + +
i)
x2 − x + 1 = (x + −
2
2
2
2
So
g(x) = x6 − 1 = (x + 1)(x2 − x + 1)(x − 1)(x2 + x + 1)
x3 + 1 = (x + 1)(x2 − x + 1).
= (x + 1)(x +
So we now have
1
2
√
−
3
2 i)(x
+
1
2
√
+
3
2 i)(x
− 1)(x −
1
2
√
−
3
2 i)(x
−
1
2
√
+
3
2 i)
g(x) = x6 − 1 = (x3 + 1)(x3 − 1) = (x + 1)(x2 − x + 1)(x − 1)(x2 + x + 1).
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We still
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Complex Numbers
In the next presentation, we return to studying the zeroes of polynomials.
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Notice that in this last example, relying on complex numbers and the
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formula for difference of squares, we were able to break the sixth degree
polynomial x6 − 1 down into six linear terms as promised by the
Fundamental Theorem of Algebra.
.