P3 Chapter 16 CIE Centre A-level Pure Maths © Adam Gibson.
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Transcript of P3 Chapter 16 CIE Centre A-level Pure Maths © Adam Gibson.
![Page 1: P3 Chapter 16 CIE Centre A-level Pure Maths © Adam Gibson.](https://reader030.fdocuments.us/reader030/viewer/2022020710/5519cf20550346443e8b49c1/html5/thumbnails/1.jpg)
P3 Chapter P3 Chapter 1616
CIE Centre A-level CIE Centre A-level Pure MathsPure Maths
© Adam Gibson
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COMPLEX NUMBERSCOMPLEX NUMBERS
Girolamo Cardano: 1501-1576
A colourful life!
3 0x px q
The “depressed cubic”;other cubics can be expressedin this form.
3 2ax bx cx d
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COMPLEX NUMBERSCOMPLEX NUMBERS
Cardano “stole” the method from Tartaglia. Finding x you mustwrite:
2 3
3 2 , 2 4 27 3 3
q q p p au x u
u
can be less than zerobut the solutionis a real number!
Square roots of negative numbers can be useful,just as negative numbers themselves.
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COMPLEX NUMBERSCOMPLEX NUMBERS
By introducing one number 2 1i
we can solve lots of new problems, and makeother problems easier.
It can be multiplied, divided, added etc. just as anyother number; but the equation above is the onlyextra rule that allows you to convert between i andreal numbers.
Start by noticing that
4 4 1 2i So the square root ofany negative numbercan be expressed in termsof i.
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COMPLEX NUMBERSCOMPLEX NUMBERS
Therefore we can solve equations like:
2 8x The answer is 2 2x i
But what is 1 3i 1 3i
The two types of number cannot be “mixed”.
Numbers of the form , k i k are called imaginary numbers (or “pure imaginary”)
Numbers like 1, 2, -3.8 that we used before are calledreal numbers.
When we combine them together in a sum we havecomplex numbers.
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COMPLEX NUMBERSCOMPLEX NUMBERS
To summarize,
z a bi
•a and b are real numbers•a is the “real part” of z; Re(z) •b is the “imaginary part” of z; Im(z)•The sum of the two parts is called a “complex number”
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COMPLEX NUMBERSCOMPLEX NUMBERS
Adding and subtracting complex numbers:
1
2
(2 3 )
(4 9 )
z i
z i
1 2z z 6 6i
( ) ( ) ( ) ( )a bi c di a c b d i
For addition and subtraction the real and imaginaryparts are kept separate.
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COMPLEX NUMBERSCOMPLEX NUMBERS
Multiplying and dividing complex numbers:
1
2
(2 3 )
(4 9 )
z i
z i
1 2z z
2
(2 3 ) (4 9 )
2 4 (2 9 ) (3 4) (3 9 )
8 18 12 ( 27 )
35 6
i i
i i i i
i i i
i
( ) ( ) ( ) ( )a bi c di ac bd bc ad i
Notice how, for multiplication, the real and imaginaryparts “mix” through the formula i2 = -1.
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COMPLEX NUMBERSCOMPLEX NUMBERS
Multiplying and dividing complex numbers:
1
2
(2 3 )
(4 9 )
z i
z i
1
2
z
z (2 3 )
(4 9 )
i
i
(2 3 ) (4 9 )
(4 9 ) (4 9 )
i i
i i
2
2
8 18 12 (27 )
4 4 36 36 ( 9 9 )
i i i
i i i
19 30 19 30
97 97 97
ii
Read through Sections
16.1 and 16.2 to makesure you understand the basics.
Rememberthis trick!!
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COMPLEX CONJUGATESCOMPLEX CONJUGATES
Now that we have introduced complex numbers, we can view the quadratic solution differently.
2 4
2
b b acx
a
Now there are always two solutions, albeit they can berepeated real solutions.
If the equation has no real roots, it must have two complex roots.If one complex root is 1 8i what is the other?
1 8i These two numbers are called“complex conjugates”.
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COMPLEX CONJUGATESCOMPLEX CONJUGATES
What are the solutions to 2 6 21 0x x ?
3 2 3i
If we write 3 2 3z i
Then the complex conjugate is written as * 3 2 3z i
* means conjugate
Calculate the following:
*
*
*
z z
z z
zz
22
6 2Re( )
4 3 2 Im( )
3 2 3 21
z
i z
2z
This will be discussedlater.
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COMPLEX POWERSCOMPLEX POWERS
What happens if we square a complex number z?
2
2 2
2 2
( )( )
( ) (2 )
z x iy
z x iy x iy
x xiy xiy y
x y i xy
*
2*
2 2
( )( )
(2 )
z x iy
z x iy x iy
x y i xy
Compare the two results;they are complex conjugates!
And then squareits conjugate, z*:
* 22 *z z Later we will understandthis result geometrically
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COMPLEX POWERSCOMPLEX POWERS
Continuing these investigations further (you may studyin your own time if you wish):
* *
* * *1 2 1 2
nnz z
z z z z
This will be easy to justifylater.
Examine the argument on page 228. This is a key idea, although you don’t have to understand the proof.
Non-real roots of polynomialsNon-real roots of polynomials with real coefficientswith real coefficients
always occur in conjugate pairs.always occur in conjugate pairs.
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COMPLEX POWERSCOMPLEX POWERS
Tasks
Find all the roots of the following two polynomials:
4 2b) 1z z
3 2a) 7 65z z z
The first example can be attacked using the factortheorem.Examining +/-1,+/-5,+/-13 gives one root as -5.Equating coefficients therefore gives:
3 2 27 65 ( 5)( 4 13)z z z z z z
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COMPLEX POWERSCOMPLEX POWERS
3 2 7 65 ( 5)( 2 3 )( 2 3 )z z z z z i z i
The roots are therefore -5, 2-3i, 2+3i.
4 2b) 1z z
The second example looks simpler but is, in a way, more difficult.First set w = z2. 2
2 2
1 0
1 3
21 3 1 3
or 2 2 2 2
w w
w
i iz z
It seems we haveto find the squareroot of a complexnumber!
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COMPLEX POWERSCOMPLEX POWERS
Algebra is not the best way to do it, but let’s try anyway.
( )( )
z a ib
a ib a ib z
The next step is important tounderstand. It is called“equating real and imaginary parts”.
2 2
Re :
Im :
2
a b x
ab y
22
2
4
ya
by
b xb
Simultaneousequations.Let’s apply it to our problem.
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COMPLEX POWERSCOMPLEX POWERS
2
2 2
22
4 2
2
1 3
2 21 3
( )( )2 2
1 3 , 2
2 29 1
16 2
16 8 9 0
8 24 1 3 11 or
32 4 4 2
iz
ix iy x iy
x y xy
xx
x x
x
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COMPLEX CONJUGATESCOMPLEX CONJUGATES
Special properties of complex conjugates:
z
z*
* 2Re( )z z z * 2Im( )z z z
What is *zz ?
How do we knowit must be a realnumber?
Im( )z
Re( )z
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COMPLEX CONJUGATESCOMPLEX CONJUGATES
*
(cos sin )
(cos sin )
z r i
z r i
*
2* * 2
arg( ) 0
or
zz
zz z z z r
2*zz z This is a very important result.
1 2 1 2arg( ) arg( ) arg( )z z z z
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COMPLEX CONJUGATESCOMPLEX CONJUGATES
How many complex roots do the following polynomialshave?
10
2 3
2 5 4
A 3 4
B 65 63
C 3 4 18 13
z z
z z z
z z z
10
3
5
See page 229. We always have n roots for a polynomial of degree n. If the coefficients are real numbers, then we also know that any non-real roots occur in complex conjugate pairs.
If 1-8i is a root of polynomial B, what are the other roots?
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POLAR COORDINATE FORMPOLAR COORDINATE FORM
zIm( )z
Re( )z
r
cosr
sinr
The modulus is the length of the line from 0+0i to the number z, i.e. r.The argument is the angle between the positive real axis and that line, by convention we use
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POLAR COORDINATE FORMPOLAR COORDINATE FORM
Find, to 3 s.f. the modulus and argument of the followingcomplex numbers:
4
7 6
i
i
arg( )z 2 24 1 17r
(cos sin )z r i
To find θ we have two equations:
4 1cos , sin
17 17
0.245
modulus
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POLAR COORDINATE FORMPOLAR COORDINATE FORM
4
7 6
i
i
2 26 7 17 5 9.22 (3 s.f.)z r
7 6cos , sin
17 5 17 5
arg( ) 2.43 (3 s.f.)z
7 6 9.22(cos(2.43) sin(2.43))z i i
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EXPONENTIAL FORMEXPONENTIAL FORM
( ) cos( ) sin( )y x x i x
or
( ) cos( ) sin( )z i
dy
dx sin cosx i x ( sin cos )i i x x
iy
Which function does this?dy
kydx
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EXPONENTIAL FORMEXPONENTIAL FORM
kxy Ae
So (not proof but good enough!)
(cos sin ) iz r i re
If you find this incredible or bizarre, it means youare paying attention.
Substituting gives “Euler’s jewel”:
1 0ie which connects, simply, the 5 mostimportant numbers in mathematics.
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EXPONENTIAL FORMEXPONENTIAL FORM
We can write any complex number in this form ire
As before, r is the modulus and θ is the argument.
Examples:
2
6
1 3
i
i
i
2
2
3
0
2
6
2
i
i
i
i
e
e
e
e
Do you see how easy itis to calculate powers?
Find 10
1 3i
23
1
1024ie