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Complex Numbers Summary
Academic Skills Advice
What does a complex number mean?
A complex number has a ‘real’ part and an ‘imaginary’ part (the imaginary part involves the
square root of a negative number).
e.g. Z = a + π’Šb
We use Z to denote a complex number:
Real
Example:
Imaginary
You might see the 𝑖 before or after it’s
number - it doesn’t matter which.
Z = 4 + 3i
Re(Z) = 4 Im(Z) = 3
Sometimes (especially in engineering)
a j is used instead of 𝑖 – they mean the
same thing.
Powers of i
𝑖 stands for √−1
so:
For any power of 𝑖 take out as many 𝑖
and they will all end up as ±π‘– or ±1.
Example:
2
𝑖 2 = (√−1) = -1
𝑖 4 = (𝑖 2)2 = (-1)2 = 1
4’s
and 𝑖
2’s
as possible
Summary:
π’Š = √−𝟏
π’ŠπŸ = −𝟏
π’ŠπŸ’ = 𝟏
π’Š−𝟐 = −𝟏
π’Š−πŸ’ = 𝟏
𝑖 11 = (𝑖 4 )2 𝑖 2 𝑖 = 12 × −1 × π‘– = −𝑖
Adding & Subtracting
This is easy – just add or subtract the real part and add or subtract the imaginary parts:
Examples:
(4 + 3𝑖 ) + (2 + 6𝑖 ) = (6 + 9𝑖 )
(3 + 7𝑖 ) – (1 – 3𝑖 ) = (2 + 10𝑖 )
Multiplying
Multiply out the 2 brackets.
Example:
(3 + 5𝑖 )(4 – 2𝑖 ) = 12 – 6𝑖 +20𝑖 – 10𝑖 2 = 12 + 14𝑖 – 10 (-1) = 22 + 14𝑖
Complex Conjugate
The conjugate is exactly the same as the complex number but with the opposite sign in the
middle. When multiplied together they always produce a real number because the middle
terms disappear (like the difference of 2 squares with quadratics).
Example:
(4 + 6𝑖 )(4 – 6𝑖 ) = 16 – 24𝑖 + 24𝑖 – 36𝑖 2 = 16 – 36(-1) = 16 + 36 = 52
© H Jackson 2010 / Academic Skills
1
Dividing
Dividing by a real number:
divide the real part and divide the imaginary part.
Dividing by a complex number:
Multiply top and bottom of the fraction by the complex
conjugate of the denominator so that it becomes real, then
do as above.
Examples:
3+4𝑖
2
4−5𝑖
3+2𝑖
3
=
2
4
+ 𝑖 = 1.5 + 2𝑖
2
4−5𝑖
=
3+2𝑖
×
3−2𝑖
12−8𝑖−15𝑖+10𝑖 2
=
3−2𝑖
9−6𝑖+6𝑖−4𝑖 2
=
12−23𝑖+10(−1)
9−4(−1)
=
2−23𝑖
13
=
𝟐
πŸπŸ‘
−
πŸπŸ‘
πŸπŸ‘
π’Š
Graphical Representation
A complex number can be represented on an Argand diagram by plotting the real part on
the π‘₯-axis and the imaginary part on the y-axis.
Example:
Imaginary
P (𝑧
= π‘₯ + 𝑦𝑖)
y
0
Nb Tan(θ) =
𝑦
π‘₯
πœƒ
π‘₯
Real
is written as |𝒛| and is the length of OP, therefore |𝑧| = √π‘₯ 2 + 𝑦 2
is the angle θ that is made with the horizontal axis (denoted by ∠).
Modulus:
Argument:
Properties of the Modulus and Argument
𝑧1 =
Let
𝑧2
Then |𝑧1 | =
These let you find
the modulus and the
argument when one
complex number is
divided by another.
πœ‹
Examples:
|𝑧2 |
Then |𝑧1 | = |𝑧2 | × |𝑧3 |
|𝑧3 |
∠𝑧1 = ∠𝑧2 − ∠𝑧3
And
πœ‹
And
𝑧
if 𝑧2 = 5𝑒 2 𝑖 , 𝑧3 = 3𝑒 3 𝑖 , 𝑧1 = 𝑧2
3
Then |𝑧1 | =
And
|𝑧2 |
|𝑧3 |
5
∠𝑧1 = ∠𝑧2 − ∠𝑧3
πœ‹
πœ‹
5
πœ‹
∴ 𝑧1 = 3 𝑒 6 𝑖
© H Jackson 2010 / Academic Skills
∠𝑧1 = ∠𝑧2 + ∠𝑧3
πœ‹
if
These let you find the
modulus and the
argument when one
complex number is
multiplied by another.
πœ‹
𝑧2 = 5𝑒 2 𝑖 , 𝑧3 = 3𝑒 3 𝑖 , 𝑧1 = 𝑧2 𝑧3
Then |𝑧1 | = |𝑧2 | × |𝑧3 | = 15
=3
=2−3 =
𝑧1 = 𝑧2 𝑧3
Let
𝑧3
πœ‹
6
And
∠𝑧1 = ∠𝑧2 + ∠𝑧3
πœ‹
πœ‹
=2+3
=
5πœ‹
6
5πœ‹
∴ 𝑧1 = 15𝑒 6 𝑖
2
Polar Form
Once you have shown a complex number on an Argand diagram then it’s easy to find the
length of the line and the angle it makes with the π‘₯-axis.
To find the length: use Pythagoras
To find the angle: use tan 𝜽
Once you have this info you can write the complex number in polar form:
𝒛 = π’“π’„π’π’”πœ½ + π’Šπ’“π’”π’Šπ’πœ½
𝒛 = 𝒓(π’„π’π’”πœ½ + π’Šπ’”π’Šπ’πœ½)
Example:
Express 𝑧 = 3 + 4𝑖 in polar form
imaginary
Pythagoras: π‘Ÿ = √32 + 42 = √16 + 9 = √25 = 5
π‘Ÿ
0
Where π‘Ÿ is the length of the line and πœƒ
is the angle it makes with the π‘₯-axis.
4
πœƒ
3
4
πœƒ = 53.1o
Tan πœƒ:
tanπœƒ = 3
Polar form:
𝑧 = 5(cos 53.1 + 𝑖𝑠𝑖𝑛53.1)
real
Nb if, when you draw the complex number, it’s in a different quadrant adjust the angle as necessary.
Exponential Form
𝒛 = π’“π’†π’Šπœ½
The 3 forms:
(where π‘Ÿ and πœƒ are the same as in polar form but πœƒ must be in radians)
Basic
𝒛 = 𝒂 + π’Šπ’ƒ
Polar
𝒛 = 𝒓(π’„π’π’”πœ½ + π’Šπ’”π’Šπ’πœ½)
Exponential
𝒛 = π’“π’†π’Šπœ½
De Moivres Theorem:
Is used for raising a complex number to a power.
𝒛𝒏 = 𝒓𝒏 (π’„π’π’”π’πœ½ + π’Šπ’”π’Šπ’π’πœ½)
πœ‹
e.g
e.g.2: (1 + 𝑖)100
If 𝑧 = 3𝑒 3 𝑖
then 𝑧 5 = 35 (π‘π‘œπ‘ 
© H Jackson 2010 / Academic Skills
5πœ‹
3
+ 𝑖𝑠𝑖𝑛
5πœ‹
)
3
We could use De Moivres or:
(1 + 𝑖)100 = ((1 + 𝑖)2 )50
3
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