Upload
others
View
1
Download
0
Embed Size (px)
Citation preview
Imaginary Number and the Symbol ๐ฝ
2
Consider the solutions for this quadratic equation: ๐ฅ2 + 1 = 0
๐ฅ = ยฑ โ1
โ1 is called the imaginary number, and we use the symbol ๐ to
represent it: ๐ = โ1.
Thus, the solutions ๐ฅ = ยฑ๐
Obviously, ๐2 = โ1
This notation allowed us to deal with a large number of quadratic equations of
the form: ๐๐ฅ2 + ๐๐ฅ + ๐ = 0
with the solution ๐ฅ =โ๐ยฑ ๐2โ4๐๐
2๐
In case ๐2 โ 4๐๐ < 0, so we can rewrite
๐ฅ =โ๐ ยฑ ๐ 4๐๐ โ ๐2
2๐
Complex Numbers
3
Example: ๐ฅ2 + 2๐ฅ + 3 = 0
We have two roots
๐ฅ1 = โ1 + ๐ 2 and ๐ฅ2 = โ1 โ ๐ 2
Both roots consisting of a real number and an imaginary number are
complex number.
Re(๐ฅ1) = โ1 โ real part of the complex number ๐ฅ1
Im(๐ฅ1) = 2 โ imaginary part of the complex number ๐ฅ1
Re(๐ฅ2) = โ1 โ real part of the complex number ๐ฅ2
Im ๐ฅ2 = โ 2 โ imaginary part of the complex number ๐ฅ2
Note: Both Re(๐ฅ) and Im(๐ฅ) are real numbers themselves.
Operations of Complex Numbers
4
Addition: โ2 + ๐6 + 3 โ ๐4 = 1 + ๐2
Subtraction: โ2 + ๐6 โ 3 โ ๐4 = โ5 + ๐10
Powers of ๐,
๐2 = โ1, ๐3 = ๐2๐ = โ๐, ๐4 = ๐2๐2 = โ1 โ1 = 1, โฆ
Multiplication:
โ2 + ๐6 3 โ ๐4 = โ2 3 + โ2 โ๐4 + ๐6 3 + (๐6)(โ๐4)
= โ6 + ๐8 + ๐18 โ ๐224 = โ6 + ๐26 โ โ1 24= 18 + ๐26
Complex Conjugates
Complex conjugate:
๐ + ๐๐ and ๐ โ ๐๐ are complex conjugate pairs
๐ + ๐๐ ๐ โ ๐๐ = ๐2 โ ๐๐ 2 = ๐2 + ๐2
โ always a real number
Example:
5
3 + ๐4 3 โ ๐4 = 32 + 42 = 25
Division of Complex Numbers
6
Division by a real number 3+๐4
2= 1.5 + ๐2
Division by another complex number
3 + ๐4
2 โ ๐3=
3 + ๐4 2 + ๐3
2 โ ๐3 2 + ๐3
=6 โ 12 + ๐(8 + 9)
4 + 9=
โ6 + ๐17
13= โ0.46 + ๐1.31
Division by itself
3 + ๐4
3 + ๐4=
3 + ๐4 3 โ ๐4
3 + ๐4 3 โ ๐4=
25
25= 1
Equal Complex Numbers
7
If two complex number are equal, their corresponding real and imaginary
parts must be equal:
๐ + ๐๐ = ๐ + ๐๐
Implies:
๐ = ๐, and ๐ = ๐
Because we can write
๐ โ ๐ = ๐(๐ โ ๐)
And a real number never equals an imaginary number.
Example:
๐ฅ = ๐ฅ1 + ๐๐ฅ2 = 6 + ๐8
Then ๐ฅ1 = 6, and ๐ฅ2 = 8.
Complex Plane (Argand Diagram)
8
Real axis
Imaginary axis ๐ง = ๐ + ๐๐
(๐, ๐)
๐
๐
Graphical addition of two complex numbers
2
2
3
1
3
5
Polar Form
9
Real axis
Imaginary axis
๐ง = ๐ + ๐๐
๐
๐ ฮธ
๐
Conversion between polar and rectangular forms
๐ = ๐2 + ๐2, ๐ = tanโ1๐
๐
๐ = ๐ cos ๐ , ๐ = ๐ sin ๐ , โ๐ โค ๐ โค ๐
Example: ๐ง = 6 โ ๐8 can be converted into polar form,
๐ = 62 + 82 = 10, ๐ = tanโ1 โ8
6= โ36๐52โฒ
๐: modulus of the complex number ๐ง
๐: argument of the complex number ๐ง
Exponential Form
10
Maclaurin Series
๐๐ฅ = 1 + ๐ฅ +๐ฅ2
2!+
๐ฅ3
3!+
๐ฅ4
4!+
๐ฅ5
5!+ โฏ
Let ๐ฅ = ๐๐,
๐๐๐ = 1 + ๐๐ +๐๐ 2
2!+
๐๐ 3
3!+
๐๐ 4
4!+
๐๐ 5
5!+ โฏ
= 1 โ
๐2
2!+
๐4
4!โ โฏ + ๐ ๐ โ
๐3
3!+
๐5
5!โ โฏ
= cos ๐ + ๐ sin ๐ โ ๐๐ฎ๐ฅ๐๐ซโฒ๐ฌ ๐ข๐๐๐ง๐ญ๐ข๐ญ๐ฒ
Thus, ๐ cos ๐ + ๐ sin ๐ = ๐๐๐๐ โ exponential form
Three ways: ๐ง = ๐ + ๐๐ = ๐ cos ๐ + ๐ sin ๐ = ๐๐๐๐
Which Form Is Better?
11
It depends on what you want to do:
For addition and subtraction, it is better to work with the
rectangular form ๐ง = ๐ + ๐๐
For multiplication and division, it is better to work with
exponential form, ๐ง = ๐๐๐๐, e.g.,
4๐๐60๐
2๐๐30๐ = 2๐๐ 60๐โ30๐= 2๐๐30๐
To convert between rectangular and exponential form, we work
thru polar form, ๐ง = ๐ cos ๐ + ๐ sin ๐
3 + ๐4 = 5 cos 36๐52โฒ + ๐ sin 36๐52โฒ = 5๐๐36๐52โฒ
Powers
12
Consider: ๐ง = ๐๐๐๐
๐ง2 = ๐๐๐๐ 2= ๐2๐๐2๐ = ๐2 cos 2๐ + ๐ sin 2๐ , โฆ ๐ง๐ = ๐๐๐๐ ๐
=
๐๐๐๐๐๐ = ๐๐ cos ๐๐ + ๐ sin ๐๐ โ DeMoivreโs theorem
If ๐๐ > ๐, we need to add or subtract multiples of 2๐ so that the equivalent
angle ๐ stays within ๐ โค ๐.
Example: ๐ง = 2๐๐120๐= 2โ 120๐, ๐ง5 = 25๐๐5ร120๐
= 32โ 600๐ = 32โ โ120๐
120๐
โ120๐
๐ง
๐ง5
Roots
13
Roots of a complex number is somewhat complicated and requires careful
attention in manipulation.
Example, ๐ค = 32โ โ120๐= 32๐โ๐120๐, letโs find ๐ค1/5=?
1) We can take do the following
32๐โ๐120๐ 1/5= 32 1/5๐โ๐120๐/5 = 2๐โ๐24๐
= ๐โ โ ๐๐๐
But we certainly expect to recover ๐ง = 2๐๐120๐= 2โ 120๐ because
๐ง5 = 32โ โ120๐ from the previous example of powers, so what
happened?
2) Of course, we can say โ โ 120๐ is also โ 360๐ โ 120๐ = โ 240๐, so now
32๐โ๐120๐ 1/5= 32๐๐240๐ 1/5
= 32 1/5๐๐240๐/5 = 2๐๐48๐= ๐โ ๐๐๐
But the solution ๐ง = 2๐๐120๐= 2โ 120๐ is still not recovered.
Roots (Contโd)
14
3) Well, we can keep adding โ 720๐ โ 120๐ = โ 600๐, so now
32๐โ๐120๐ 1/5= 32๐๐600๐ 1/5
= 32 1/5๐๐600๐/5 = 2๐๐120๐= ๐โ ๐๐๐๐ This is
it, but how do we know we have gotten them all?
4) We can also subtract 360๐, so โ โ 360๐ โ 120๐ = โ โ480๐, then
32๐โ๐120๐ 1/5= 32๐โ๐480๐ 1/5
= 32 1/5๐โ๐480๐/5 = 2๐โ๐96๐= ๐โ โ๐๐๐
5) Of course, we can also obtain
32๐โ๐120๐ 1/5= 32๐โ๐840๐ 1/5
= 32 1/5๐โ๐840๐/5 = 2๐โ๐168๐= ๐โ โ๐๐๐๐
When do we stop?
Roots (Contโd)
15
Letโs plot these roots in the complex plane.
๐โ โ ๐๐๐
๐โ ๐๐๐ ๐โ ๐๐๐๐
๐โ โ๐๐๐
๐โ โ๐๐๐๐
๐๐๐โ๐๐๐๐๐
We can see that these five roots are
equally separated by 360๐
5= 72๐.
It is generally true that the n-th order roots have n number of values that are
equally separated by ๐๐๐๐
๐.
So in practice, we only need to find one as we did in Step 1), and the remaining ๐ โ 1 roots will be generated similar to the illustration in the complex plane plot.
The one nearest to the positive real axis is called the principal root.
1
2
3
4 5
Summary Key points:
Recognize ๐ = โ1 for imaginary number
Three forms of the complex number: rectangular,
polar, and exponential
Conversions between the three forms
Complex conjugate pairs
Addition, subtraction, multiplication (powers), and
division
Roots of ๐-th order (equally spaced by 360๐/๐ on a
circle)
16