Decision Maths - Haringeymath's Blog
Download
Report
Transcript Decision Maths - Haringeymath's Blog
Further Pure 1
Lesson 5 – Complex
Numbers
Numbers
Wiltshire
What types of numbers do we already know?
Real numbers – All numbers ( 2, 3.15, π ,√2)
Rational – Any number that can be expressed as a
fraction
– (4, 2.5, 1/3)
Irrational – Any number that can`t be expressed as a
fraction. ( π ,√2, √3 + 1)
Natural numbers – The counting numbers (1, 2, 3, ….)
Integers – All whole numbers ( -5, -1, 6)
Complex Numbers (Imaginary numbers)
Complex Numbers
Wiltshire
What is the √(-1)?
We define the √(-1) to be the imaginary
number j. (Hence j2 = -1)
Note that lots of other courses use the letter
i, but we are going to use j.
We can now use this to calculate a whole
new range of square roots.
What is √(-144)?
Answer is +12j and -12j, or ±12j.
Complex Numbers
Wiltshire
Now we can define a complex number (z) to
be a number that is made up of real and
imaginary parts.
z=x+yj
Here x and y are real numbers.
x is said to be the real part of z, or Re(z).
y is said to be the imaginary part of z, or
Im(z).
Solving Quadratics
Wiltshire
Use the knowledge you have gained in the last few
slides to solve the quadratic equation
z2 + 6z + 25 = 0
Remember
Solution
b b 2 4ac
z
2a
6 62 4 1 25
z
2 1
Solving Quadratics
6 36 100
z
2
6 64
z
2
68 j
z
2
z 3 8 j
Wiltshire
Addition and Subtraction
Wiltshire
To Add and subtract complex numbers all you
have to do is add/subtract the real and
imaginary parts of the number.
(x1+y1j) + (x2+y2j)
= x 1 + x 2 + y 1j + y 2j
= (x1 + x2)+ (y1 + y2)j
(x1+y1j) – (x2+y2j)
= x 1 - x 2 + y 1j - y 2j
= (x1 - x2)+ (y1 - y2)j
Multiplying
Wiltshire
Multiplying two complex numbers is just like
multiplying out two brackets.
You can use the FOIL method.
First Outside Inside Last.
Remember j2 = -1
(x1+y1j)(x2+y2j) = x1x2 + x1y2j + x2y1j + y1y2j2
= x 1x 2 + x 1y 2j + x 2y 1j - y 1y 2
= x 1x 2 - y 1y 2 + x 1y 2j + x 2y 1j
= (x1x2 - y1y2) + (x1y2 + x2y1)j
What is j3, j4, j5?
Multiplying
Wiltshire
Alternatively you could use the box method.
x1
y1j
x2
x1x2
x2y1j
y2j
x1y2j - y1y2
(x1+y1j)(x2+y2j) = (x1x2 - y1y2) + (x1y2 + x2y1)j
Questions
If z1 = 5 + 4j
z2 = 3 + j
Find
a)
z1 + z3 = 12 + 2j
b)
z1 - z2 = 3 + 3j
c)
z1 – z3 = -2 + 6j
d)
z1 × z2 = 11 + 17j
e)
z1 × z3 = 43 + 18j
Wiltshire
z3 = 7 – 2j
Complex Conjugates
Wiltshire
The complex conjugate of
z = (x + yj) is z* = (x – yj)
If you remember the two solutions to the quadratic
from a few slides back then they where complex
conjugates.
z = -3 + 8j & z = -3 – 8j
In fact all complex solutions to quadratics will be
complex conjugates.
If z = 5 + 4j
What is z + z*
What is z × z*
Activity
Wiltshire
Prove that for any complex number z = x + yj,
that z + z* and z × z* are real numbers.
First
z + z* = (x + yj) + (x – yj)
= x + x + yj – yj
= 2x = Real
Now
z × z* = (x + yj)(x – yj)
= x2 – xyj + xyj – y2j2
= x2 – y2(-1)
= x2 + y2 = Real
Now complete Ex 2A pg 50
Division
Wiltshire
There are two ways two solve problems involving
division with complex numbers.
First you need to know that if two complex numbers
are equal then the real parts are identical and so are
there imaginary parts.
If we want to solve a question like 1 ÷ (2 + 4j) we first
write it equal to a complex number p + qj.
1
p qj
24 j
Now we re-arrange the equation to find p and q.
(p + qj)(2 + 4j) = 1
Division
Wiltshire
Expanding the equation gives
2p – 4q + 2qj + 4pj = 1
The number 1 can be written as 1 + 0j
So
(2p – 4q) + (2q + 4p)j = 1 + 0j
Now we can equate real and imaginary parts.
2p – 4q = 1
4p + 2q = 0
Solve these equations
p = 1/10
&
q = -1/5
Therefore 1 ÷ (2 + 4j) = 0.1 – 0.2j
Division
Wiltshire
The second method is similar to rationalising the
denominator in C1.
1
1
24 j 24 j 1 1
j
24 j 24 j 24 j
20
10 5
The 20 on the bottom comes from the algebra we
proved a few slides back. (x + yj)(x – yj) = x2 + y2
Now see if you can find (3 - 5j) ÷ (2+9j)
3 5 j 3 5 j 2 9 j 6 10 j 27 j 45 39 37 j
29 j 29 j 29 j
4 81
85
Now complete Ex 2B pg 53
Argand Diagrams
Complex numbers can
be shown Geometrically
on an Argand diagram
The real part of the
number is represented
on the x-axis and the
imaginary part on the y.
-3
-4j
3 + 2j
2 – 2j
Wiltshire
Im
Re
Modulus of a complex number
Wiltshire
A complex number
can be represented
by the position vector.
Im
x
y
The Modulus of a
complex number is
the distance from the
origin to the point.
|z| = √(x2+y2)
Note |x| = x
y
x
Re
Modulus of a complex number
Wiltshire
Find
a)
|3 + 4j| = 5
b)
|5 - 12j| = 13
c)
|6 - 8j| = 10
d)
|-24 - 10j| = 26
Sum of complex numbers
z1 + z
2
=
Im
x1 x2 x1 x2
y1 y2 y1 y2
2 5 2 5 7
5 1 5 1 6
Wiltshire
z2
z1
z1 + z2
Re
Difference of complex numbers
Wiltshire
z2 - z1 =
Im
x2 x1 x2 x1
y2 y1 y2 y1
z1
7 2 7 2 5
6 5 6 5 1
z2
z2 – z1
Now complete
Ex 2C pg 57
Re
Sets of points in Argand diagram
Wiltshire
What does |z2 – z1|
represent?
If z1 = x1 +y1j
& z2 = x2 +y2j
Then z2 – z1
= (x2 – x1) + (y2 – y1)j
So |z2 – z1|
= √((x2 – x1)2 + (y2 – y1)2)
This represents the
distance between to
complex numbers
z1 & z2.
Im
(x2,y2)
y2- y1
(x1,y1)
x2- x1
Re
Examples
Draw an argand diagram
showing the set of points
for which |z – 3 – 4j| = 5
Solution
First re-arrange the
question
|z – (3 + 4j)| = 5
From the previous slide this
represents a constant
distance of 5 between the
point (3,4) and z.
This will give a circle centre
(3,4)
Now do Ex 2D pg 60
Wiltshire
Im
Re
Examples
Wiltshire
How would you show the sets of points for
which:
i) |z – 3 – 4j)| ≤ 5
ii) |z – 3 – 4j)| < 5
iii) |z – 3 – 4j)| ≥ 5
Im
Im
Im
Re
Re
Re