#### Transcript Decision Maths - Haringeymath's Blog

```Further Pure 1
Lesson 5 – Complex
Numbers
Numbers
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 What types of numbers do we already know?
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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
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 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
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 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).
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 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
 6  36  100
z
2
 6   64
z
2
68 j
z
2
z  3  8 j
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 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
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 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
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 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
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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
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z3 = 7 – 2j
Complex Conjugates
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 The complex conjugate of
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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
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 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
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 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 
24 j
 Now we re-arrange the equation to find p and q.
(p + qj)(2 + 4j) = 1
Division
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 Expanding the equation gives
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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
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 The second method is similar to rationalising the
denominator in C1.
1
1
24 j 24 j 1 1



  j
24 j 24 j 24 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


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29 j 29 j 29 j
4  81
85
 Now complete Ex 2B pg 53
Argand Diagrams
 Complex numbers can
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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
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Im
Re
Modulus of a complex number
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 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
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 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
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z2
z1
z1 + z2
Re
Difference of complex numbers
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 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
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 What does |z2 – z1|
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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
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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
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Im
Re
Examples
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 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
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