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

```Further Pure 1
Lesson 6 – Complex
Numbers
Modulus/Argument
of a complex number

The position of a complex
number (z) can be
represented by the
distance that z is from the
origin (r) and the angle
axis (θ).
Distance is given by r = |z|

r = |z| = √(x2+y2)

r is known as the
modulus of a complex
number.

Wiltshire
Im
r
y
θ
x
Re
Argument of a
complex number
Now lets look at the angle
(θ) that the line makes with
the positive real axis.
 NOTE: θ is measured in
direction from the positive
real axis.
 θ is measured from –π to π
and is known as the
principal argument of z.
 Argument z = arg z = θ
 y = r sinθ, x = r cosθ
 tan θ = y/x
 θ = inv tan (y/x)

Wiltshire
Im
r
y
θ
x
Re
Modulus/Argument
of a complex number
Wiltshire
 Calculate the modulus and argument of the following complex
numbers. (Hint, it helps to draw a diagram)
1)
3 + 4j
|z| = √(32+42) = 5
arg z = inv tan (4/3)
= 0.927
2)
5 - 5j
|z| = √(52+52) = 5√2
arg z = inv tan (5/-5)
= -π/4
3)
-2√3 + 2j
|z| = √((2√3)2+22) = 4
arg z = inv tan (2/-2√3)
= 5π/6
Modulus/Argument
of a complex number
Wiltshire
 Note, inv tan (y/x) will return answer in
 Last example on previous slide inv tan (2/2√3) on your calculator will return, -π/6,
however the answer we want is 5π/6.
 In some circumstances you way need to add
or subtract π.
 This is why a diagram is useful.
Modulus-Argument form
of a complex number
Wiltshire
 So far we have plotted the position of a complex
number on the Argand diagram by going
horizontally on the real axis and vertically on the
imaginary.
 (This is just like plotting co-ordinates on an x,y
axis)
 However it is also possible to locate the position
of a complex number by the distance travelled
from the origin (pole), and the angle turned
through from the positive x-axis.
 (This is sometimes known as Polar co-ordinates
and can be studied in another course)
Modulus-Argument form
of a complex number
Wiltshire
 (r, θ)
 (x,y)
 cosθ = x/r, sinθ = y/r
 x = r cosθ, y = r sinθ,
Im
Im
r
y
θ
x
Re
Re
Converting
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 Converting from Cartesian to Polar
 (x,y) = [√(x2+y2),(inv tan (y/x))] = (r, θ)
 Convert the following from Cartesian to Polar
i)
(1,1) = (√2,π/4)
Im
ii) (-√3,1) = (2,5π/6)
r
iii) (-4,-4√3) = (8,-2π/3)
y
θ
x
Re
Converting
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 Converting from Polar to Cartesian
 (r, θ) = (r cosθ, r sin θ)
 Convert the following from Polar to Cartesian
i)
(4,π/3) = (2,2√3)
Im
ii) (3√2,-π/4) = (3,-3)
r
iii) (6√2,3π/4) = (-6,6)
y
θ
x
Re
Using the Calculator
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 It is possible to convert from Cartesian to Polar and




(Casio fx-83MS – other calculators may be different)
The polar coordinates (√2,π/4) equal the Cartesian coordinates (1,1)
On the calculator press
Shift
Rec(
√
2
,
π
÷
4
 The answer shown is the first Cartesian co-ordinate x.
To view the y press RCL F. To view the x again press
RCL E.
 Now try and convert back from the Cartesian to the
Polar.
)
Modulus-Argument form
of a complex number
Wiltshire
 Now we can define the Modulus-Argument
form of a complex number to be:
z = r (cosθ + j sinθ)
 Here r = |z| and θ = arg z
 When writing a complex number in
Modulus-Argument form it can be helpful to
draw a diagram.
 Remember that θ will take values between
-π and π.
Modulus-Argument form
of a complex number

i)
Wiltshire
Write the following numbers in ModulusArgument form:
5 + 12j
ii) √3 – j
iii) -4 – 4j
Solutions
i) 13(cos 1.176 + j sin 1.176)
ii) 2(cos (-π/6) + j sin (-π/6))
iii) 4√2(cos(-3π/4) + j sin(-3π/4))
Modulus-Argument form
of a complex number
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 When writing a complex number in Modulus-
Argument form it is important that it is written in the
exact format as above, not:
z = -r (cosθ - j sinθ)
 (r must be positive)
 You can use the following trig identities to ensure
that z is written in the correct form.
cos(π-θ) = -cosθ
sin(π-θ) = sinθ
cos(θ-π) = -cosθ
sin(θ-π) = -sinθ
cos(-θ) = cosθ
sin(-θ) = -sinθ
Modulus-Argument form
of a complex number
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 Example: Re-write -4(cosθ + j sinθ) in Modulus-
Argument form:
-4(cosθ + j sinθ)




= 4(-cosθ - j sinθ)
= 4(cos(θ-π) + j sin(θ-π))
This is now in Modulus-Argument form.
The Modulus is 4 and the Argument is (θ-π).
This may seem a bit confusing, however if you draw
a diagram and tackle the problems like we did a few
slides ago it makes more sense.
Now do Ex 2E pg 66
Sets of points using the
Modulus-Argument form
 What do you think





arg(z1-z2) represents?
If z1 = x1 +y1j
&
z2 = x2 +y2j
Then z2 – z1 = (x2 – x1)
+ (y2 – y1)j
Now arg(z2 – z1) = inv
tan ((y2 – y1)/(x2 – x1))
This represents the
angle between the line
from z1 to z2 and a line
parallel to the real axis.
So arg(z2 – z1) = α
Wiltshire
Im
(x2,y2)
y2- y1
α
(x1,y1)
x2- x1
Re
Questions

Wiltshire
Draw Argand diagrams showing the sets of
points z for which
i) arg z = π/3
ii) arg (z - 2) = π/3
iii) 0 ≤ arg (z – 2) ≤ π/3
Im
Now do Ex 2F page 68
π/3
π/3
Re
History of complex numbers
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 In the quadratic formula (b2 – 4ac) is known as the

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discriminant.
If this is greater than zero the quadratic will have 2
distinct solutions.
If it is equal to zero then the quadratic will have 1
repeated root.
If it is less than zero then there are no solutions.
Complex numbers where invented so that we could
solve quadratic equations whose discriminant is
negative.
This can be extended to solve equations of higher order
like cubic and quartic.
In fact complex numbers can be used to find the roots
of any polyniomial of degree n.
History of complex numbers
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 In 1629 Albert Girard stated that an nth degree

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polynomial will have n roots, complex or real. Taking
into account repeated roots.
For example the fifth order equation
(z – 2)(z-4)2(z2+9) = 0 has five roots.
2, 4(twice) 3j and -3j.
For any polynomial with real coefficients solutions will
always have complex conjugate pairs.
Many great mathematicians have tried to prove the
above.
Gauss achieved it in 1799.
Complex numbers and equations
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 Given that 1 + 2j is a root of 4z3 – 11z2 + 26z –

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15 = 0
Find the other roots.
Since real coefficients, 1 – 2j must also be a root.
Now [z – (1 + 2j)] and [z – (1 – 2j)] will be factors.
So [z – (1 + 2j)][z – (1 – 2j)]
= [(z – 1) + 2j)][(z – 1) – 2j)]
= (z – 1)2 – (2j)2
= z2 – 2z + 1 – (-4)
= z2 – 2z + 5
Complex numbers and equations
Wiltshire
 You can now try to spot the remaining factors
by comparing coefficients or you can use
polynomial division.
4z – 3
z2 – 2z + 5 4z3 – 11z2
4z3 – 8z2
– 3z2
– 3z2
26z – 15 = 0
20z
6z – 15
6z – 15
0
Hence 4z3 – 11z2 + 26z – 15 = (z2 – 2z + 5)(4z – 3)
The roots are, 1 + 2j, 1 – 2j and 3/4
+
+
+
+
Complex numbers and equations
Wiltshire
 Given that -2 + j is a root of the equation
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z4 + az3 + bz2 + 10z + 25 = 0
Find the values of a and b and solve the equation.
First you need to find the values of z4,z3 & z2 so that
you can sub them in to the equation above.
z2 = (-2 + j)(-2 + j) = 3 – 4j
z3 = (3 – 4j)(-2 + j) = -2 + 11j
Z4 = (-2 + 11j)(-2 + j) = -7 - 24j
Now
(-7 -24j) + a(-2 + 11j) + b(3 – 4j) + 10(-2 + j) +25 = 0
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