Complex Numbers - Hinchingbrooke

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Lesson Objective
Understand what Complex Number are and how they fit into the mathematical landscape.
Be able to do arithmetic with complex numbers
Solve the equations:
x2 +4x + 1 = 0
x2 +4x + 4 = 0
x2 +4x + 6 = 0
What can we say about the graph of:
y = x2 + 4x + c?
What are the conditions for each case?
Arithmetic of complex numbers:
Let z = 3 + 4j
w = 2 – 5j
Find:
a) w + z
b) z – w
c) z2
d) zw

e) 
Complex Conjugates, Uniqueness and Real and Imaginary Parts:
If z = 3 + 4j we say that z* = 3 – 4j is the Complex Conjugate of z
we say that Re(z) = 3 is the real part of z
we say that Im(z*) = -4 is the imaginary part of z
Two complex numbers are identical if the imaginary and real parts are the same;
In other words if 3 - 2j = a + bj
a must be equal to 3
b must be equal to -2
Complex Conjugates, Uniqueness and Real and Imaginary Parts:
If z = 3 + 4j we say that z* = 3 – 4j is the Complex Conjugate of z
we say that Re(z) = 3 is the real part of z
we say that Im(z*) = -4 is the imaginary part of z
Let w = 2 – 5j
Find:
a) w*
b) z + w*
c) w - z*
e) w*z *
f) (zw)*
g)
1
w∗
d) (z + w)*
1 ∗
h) ∗

Complex Conjugates, Uniqueness and Real and Imaginary Parts:
If z = 3 + 4j we say that z* = 3 – 4j is the Complex Conjugate of z
Let w = 2 – 5j
Find:
a) w*
b) z + w*
c) w - z*
e) w*z *
f) (zw)*
g)
1) If y is a complex number y = a + bj
x is a complex number x = c + dj
y + y * = 2Re(y)
y - y* = 2Im(y)
(xy)* = x*y *
(x*)* = x
1
1
2Re(y)
e)  +  ∗= 2+2
Prove that: a)
b)
c)
d)
From FP1 page 53
1
w∗
d) (z + w)*
1 ∗
h) ∗

2) Find real numbers a and b (with a>0) such that
a) (a + bj)2 = 21 + 20j
b) (a + bj)2 = -40 - 42j
3) Find real numbers ‘a’ and ‘b’ such that


+
=1−
3+
1+2
4) Find real numbers z for which z2 = 2z*
5) Solve z + jw = 13 3z – 4w = 2j
for complex numbers z and w
Complex Conjugates, Uniqueness and Real and Imaginary Parts:
If z = 3 + 4j we say that z* = 3 – 4j is the Complex Conjugate of z
Let w = 2 – 5j
Find:
a) w*
2+5j
e) w*z * 26+7j
b) z + w*
5+9j
f) (zw)*
26+7j g)
1) If y is a complex number y = a + bj
x is a complex number x = c + dj
Re(y + y *) = 2a
Im(y - y*) = 2b
(xy)* = x*y *
(x*)* = x
1
1
2Re(y)
e)  +  ∗= 2+2
Prove that: a)
b)
c)
d)
From FP1 page 53
c) w - z*
1
w∗
-1-j
(2-5j)/29
d) (z + w)* 5+j
1 ∗
h) ∗
(3-4j)/25

2) Find real numbers a and b (with a>0) such that
a) (a + bj)2 = 21 + 20j
b) (a + bj)2 = -40 - 42j
3) Find real numbers ‘a’ and ‘b’ such that


+
=1−
3+
1+2
4) Find real numbers z for which z2 = 2z*
5) Solve z + jw = 13 3z – 4w = 2j
for complex numbers z and w
Lesson Objective
Be able to display complex numbers on an Argand Diagram
Understand how to find the modulus and argument of a complex number
Consider z = 3 +4j
Re(z) =
Im(z) =

=
Arg(z) =
and w = -2 – 5j
Re(w) =
Im(w) =
 =
Arg(w) =
How can we represent the following on
the Argand diagram:
a) z + w ?
b) z – w ?
What is  −  ?
What does it represent?
To generate a mandelbrot set:
Solve the equation z2 + 1 =0
By using an iterative approach.
(Like C3 coursework)
Count the number of iterations
required to get within an
acceptable margin of the solution.
Plot the starting value on an Argand
diagram, with a colour that
corresponds to the number of steps
until convergence.
Eg
Rearrange to make z = -1/z
Choose 1+i as starting value
Keep iterating until within a small
radius of i.
1) Write down the modulus and the argument of these complex numbers:
a) 2 + 2j
b) -3 +4j
c) -3-4j
d) 3j
2) What can you say about the modulus and argument of z* compared to z?
3) Let z be a complex number
On the Argand diagram, show all the
complex numbers ‘z’, such that:
a)  =6

b) Arg(z) = 3
c)  − 3 =5
d)  − 3 − 4 =5

e) Arg(z-3-4j) = 3
f)  − 3 =  − 3
g)  − 2 =2  − 3
e) -2
Let ‘z’ be a complex number
On the Argand diagram, show all the
complex numbers ‘z’, such that:
 − 3 − 4 =5
Let ‘w’ be a complex number
On the Argand diagram, show all the
complex numbers ‘w’, such that:
 − 3 − 4 =  − 3 + 2
How does this differ from:
 − 3 − 4 >  − 3 + 2
 − 3 − 4 <  − 3 + 2
?
?
z 3  2 z  2
Note that:
z a
represents the distance from the complex number a to
the complex number z
Arg(z – a) represents the angle to the complex number z from the
complex number a measured from a line parallel to
the +ve part of the real axis
If z is a complex number x + iy then:
Re(z) = x
Im(z) = y
z* = The complex conjugate of z= x - iy
z 3j  4
z2 6
z 42 j  6
z 1  z  j
z  2  2 z 1
z2j 3z
Re( z )  Im(z )
Im(z )  0
2
2
Arg ( z  1)
3
Arg ( z  2  j ) 
Arg ( 2 z )
Arg ( z )



4
3
4

Arg ( z  1)  Arg ( z 1) 
4

Arg ( z  j )   Arg ( z j )
6
Re(z )  2
1
Re( z  * )  0
z
 z2
Re
0
 z 
Lesson Objective
Modulus and Argument Form and the beginnings of powerful arithmetic
Write down the modulus and the argument of these complex numbers:
a) a = 1 + 2j
b) b = 2j
c) c = -j
d) d = -1-j
Lesson Objective
Modulus and Argument Form and the beginnings of powerful arithmetic
Write down the modulus and the argument of these complex numbers:
a) a = 1 + 2j
b) b = 2j
c) c = -j
d) d = -1-j
Section A
Find
ab
ab2
ab3
ab4
…….
In each case plot the new complex number
on the Argand diagram and find its modulus
and argument
Generalise for abn
Section B
What can you say about the modulus and

argument of ad? What about ?
Generalised results:
Let w be a complex number in the form a + ib
Then:
The modulus of w =  = 2 + 2
The argument of w = Arg(w) = the angle that w makes with the +ve section of
the real axes (usually given between 180 and -180 degrees or
We can write w as:
w =  (∅ + jsin∅)
this is called modulus/argument form (or polar form)
When two complex numbers are multiplied together, the resulting complex number:
Will have an argument = to the sum of the two original arguments and a
modulus =to the product of the two original moduli.
Puzzle 1
The points 1 + j and 3 + 4j are two adjacent corners of a square.
Where are the other corners?
Can you solve this problem? More importantly, how can you solve it using complex
numbers?
Puzzle 2
Suppose that 1 + j and 3 + 4j were two adjacent corners of an equilateral triangle, where
would the final vertex be?
This is much easier to solve using complex numbers!
Puzzle 3
Can we generalise this method to a regular ‘n’ sided shape?
This really demonstrates just how cool complex numbers are!!!!!!!!
Lesson Objective
Understand the Fundametal Thm of Algebra be able to solve cubic/quartic equations
with Real coefficients even when some roots are complex
Write down any quadratic equation with
complex roots.
Solve it.
What do you notice about the solutions?
Is it true that (wn)*
(w*)n ?
Try it with (wn)*
How can we prove this key result?
Lesson Objective
Understand the Fundamental Thm of Algebra and be able to solve cubic/quartic
equations with Real coefficients even when some roots are complex
Starter:
Is it true that: (w*)n = (wn)* for all n ≥ 0 where w is a complex number?
Try it with (w*)3
How can we prove this key result?
Lesson Objective
Understand the Fundamental Thm of Algebra be able to solve cubic/quartic equations
with Real coefficients even when some roots are complex
Write down any quadratic equation with
complex roots?
Solve it.
What do you notice about the solutions?
Lesson Objective
Understand the Fundamental Thm of Algebra be able to solve cubic/quartic equations
with Real coefficients even when some roots are complex
Write down any quadratic equation with
complex roots?
Solve it.
What do you notice about the solutions?
The Fundamental Thm of algebra states that any
polynomial equation of degree ‘n’ will have exactly
‘n’ solutions (if you count repeated roots).
The Complex Conjugate Root Thm goes further,
and states that, as long as the coefficients are real,
the solutions will come in complex conjugate
pairs.
Proof of Fundamental Thm of Algebra
(Go to University – it took Gauss years)
Proof of the Complex Conjugate Root Thm
Consider: αzn + βzn-1 + γzn-2 + …….. + δz2 + εz + ζ = 0
Then ‘z’ is a solution to the equation
Now consider:
(αzn + βzn-1 + γzn-2 + …….. + δz2 + εz + ζ)* = (0)*
(αzn)* + (βzn-1)* + (γzn-2)* + …….. + (δz2)* + (εz)* + (ζ)* = (0)*
α*(zn)* + β* (zn-1)* + γ* (zn-2)* + …….. + δ* (z2)* + ε* (z)* + (ζ)* = 0*
α*(z*)n + β* (z*)n-1 + γ* (z*)n-2 + …….. + δ* (z*)2 + ε* (z*) + (ζ*) = 0*
α(z*)n + β (z*)n-1 + γ (z*)n-2 + …….. + δ (z*)2 + ε (z*) + ζ = 0*
Example
Shows the 2 + j is a solution to z3 – z2 -7z + 15 = 0
Hence find all the other roots.
Example
Give than z = -2 + j is a solution to z4 + az3 +bz2 +10z + 25 = 0
Find the values of a and b
Hence factorise the cubic and find all 4 solutions
Example
Give than z = -2 + j is a solution to z4 + az3 +bz2 +10z + 25 = 0
Find the values of a and b
Hence factorise the cubic and find all 4 solutions
a=2b=2
Other two solutions are 1 + 2j and 1 – 2j

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