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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 = 𝒙 + 𝒊𝒚
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:
2
𝑖 2 = (√−1) = -1
𝑖 4 = (𝑖 2)2 = (-1)2 = 1
Summary:
𝒊 = √−𝟏
𝒊𝟐 = −𝟏
𝒊𝟒 = 𝟏
For any power of 𝑖 substitute as many 𝑖 2’s as possible
and they will all end up as ±𝑖 or ±1.
Examples:
𝑖 20 = (𝑖 2 )10 = (−1)10 = 1
𝑖 33 = (𝑖 2 )16 𝑖 = (−1)16 𝑖 = 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 (and remember that 𝑖 2 = −1).
Example:
(3 + 5𝑖 )(4 – 2𝑖 )
2
= 12 – 6𝑖 + 20𝑖 – 10𝑖
= 12 + 14𝑖 – 10 (−1)
= 22 + 14𝑖
H Jackson 2010 / 2015 / 2016 / Academic Skills
1
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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:
2
(4 + 6𝑖 )(4 – 6𝑖 )
= 16– 24𝑖 + 24𝑖– 36𝑖
= 16– 36(−1)
= 16 + 36
= 52
Complex
conjugate
Complex
number
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 continue as above.
Examples:
3+4𝑖
2
4−5𝑖
3+2𝑖
3
=
2
4
+ 𝑖 = 1.5 + 2𝑖
2
4−5𝑖
=
3+2𝑖
×
3−2𝑖
3−2𝑖
=
12−8𝑖−15𝑖+10𝑖 2
9−6𝑖+6𝑖−4𝑖 2
=
12−23𝑖+10(−1)
9−4(−1)
=
2−23𝑖
13
=
𝟐
𝟏𝟑
−
𝟐𝟑
𝟏𝟑
𝒊
Graphical Representation
A complex number can be represented on an Argand diagram by plotting the real part on
the 𝑥-axis and the imaginary part on the y-axis.
* When finding 𝜃 either:
Imaginary
P (𝑧
= 𝑥 + 𝑦𝑖)
y
NB tan(θ) =
𝑦
so
𝑥
𝒚
𝜽 = 𝒕𝒂𝒏−𝟏 ( )
𝒙
0
𝜃
𝑥


do a sketch to see
where the angle is
or remember: if 𝒙 < 𝟎
(i.e. negative) add 180
(𝒐𝒓 𝝅) to your answer.
Real
Modulus: is written as |𝒛| or 𝑟 and is the length of OP, therefore |𝒛| = 𝒓 = √𝒙𝟐 + 𝒚𝟐
Argument: is the angle θ that is made with the horizontal axis (denoted by ∠).
Example:
P (𝑧
= 4 + 5𝑖)
5
|𝒛| = √𝟒𝟐 + 𝟓𝟐 = √𝟒𝟏 = 𝟔. 𝟒
𝟓
𝜽 = 𝒕𝒂𝒏−𝟏 ( ) = 𝟎. 𝟗 radians
𝟒
0
𝜃
4
H Jackson 2010 / 2015 / 2016 / Academic Skills
Real
2
Polar & Exponential Form
As well as the cartesian form (𝑧 = 𝑥 + 𝑖𝑦) there are 2 alternative ways of writing a complex
number:
𝒛 = 𝒓(𝒄𝒐𝒔𝜽 + 𝒊𝒔𝒊𝒏𝜽)
Polar:
Where:
𝑟 is the modulus (length of the line) |𝑧|
𝜃 is the argument (angle it makes with the 𝑥-axis)
Exponential: 𝒛 = 𝒓𝒆𝒊𝜽
(this should be in radians for the exponential form).
Remember:
To find the modulus (length), 𝒓:
use Pythagoras
𝒚
To find the argument (angle), 𝜽: use 𝒕𝒂𝒏−𝟏 (𝒙) *
Converting between the different forms:
Cartesian
Need to find 𝒓 and 𝜽
𝑟 = √𝑥 2 + 𝑦 2
Polar or Exponential
𝑦
𝜃 = 𝑡𝑎𝑛−1 ( )
𝑥
Polar or Exponential
Example:
0
Need to find 𝒙 and 𝒚
𝑥 = 𝑟𝑐𝑜𝑠𝜃
𝑦 = 𝑟𝑠𝑖𝑛𝜃
Cartesian
Express 𝒛 = 𝟑 + 𝟒𝒊 in polar and exponential form
imaginary
𝑟
*
4
𝜃
3
Modulus:
𝑟 = √32 + 42 = √16 + 9 = √25 = 5
Argument:
𝜃 = 𝑡𝑎𝑛−1 (3) = 53.1o
Polar form:
𝑧 = 5(cos(53.1) + 𝑖𝑠𝑖𝑛(53.1))
Exp form:
𝑧 = 5𝑒 0.927𝑖
4
(in radians 𝜃 = 0.927𝑟 )
real
NB It is advisable to do a quick sketch of the complex number and check that the angle you
calculated matches the diagram. If it’s in a different quadrant adjust the angle as necessary.
Remember that angles are positive when measured anticlockwise (
)
Example:
Express 𝒛
𝝅
= 𝟕𝒆𝒊𝟑
in cartesian form
𝜋
𝑥 = 𝑟𝑐𝑜𝑠𝜃
∴ 𝑥 = 7 cos (3 ) = 3.5
𝑦 = 𝑟𝑠𝑖𝑛𝜃
∴ 𝑦 = 7 sin ( 3 ) = 6.1
Cartesian form:
𝜋
𝑧 = 3.5 + 6.1𝑖
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A reminder of the 3 forms:
Cartesian
𝒛 = 𝒙 + 𝒊𝒚
Polar
𝒛 = 𝒓(𝒄𝒐𝒔𝜽 + 𝒊𝒔𝒊𝒏𝜽)
𝑟 = √𝑥 2 + 𝑦 2
𝑥 = 𝑟𝑐𝑜𝑠𝜃
𝑦 = 𝑟𝑠𝑖𝑛𝜃
Conversions:
Exponential
𝒛 = 𝒓𝒆𝒊𝜽
𝑦
𝜃 = 𝑡𝑎𝑛−1 ( ) *
𝑥
Multiplying with Polar or Exponential form
Let
This means, when multiplying 2 complex numbers:
𝑧1 = 𝑧2 𝑧3
Multiply the 𝑟′𝑠
Add the angles (𝜃′𝑠)
Then |𝑧1 | = |𝑧2 | × |𝑧3 |
And ∠𝑧1 = ∠𝑧2 + ∠𝑧3
𝝅
𝝅
Example:
If 𝒛𝟏 = 𝟓𝒆𝟐 𝒊 and 𝒛𝟐 = 𝟑𝒆𝟑 𝒊 find 𝒛𝟏 𝒛𝟐
New modulus:
New angle:
5 × 3 = 15
𝜋
𝜋
5𝜋
+
=
2
3
6
5𝜋
∴ 𝑧1 𝑧2 = 15𝑒 6 𝑖
Dividing with Polar or Exponential form
Let
𝑧1 =
Then |𝑧1 | =
And
𝑧2
This means, when dividing 2 complex numbers:
𝑧3
Divide the 𝑟′𝑠
Subtract the angles (𝜃′𝑠)
|𝑧2 |
|𝑧3 |
∠𝑧1 = ∠𝑧2 − ∠𝑧3
𝝅
𝝅
𝝅
𝝅
𝒛
Example:
if 𝒛𝟏 = 𝟓 (𝒄𝒐𝒔 (𝟐 ) + 𝒊𝒔𝒊𝒏 ( 𝟐 )) and 𝒛𝟐 = 𝟑 (𝒄𝒐𝒔 ( 𝟑 ) + 𝒊𝒔𝒊𝒏 ( 𝟑 )) find 𝒛𝟏
New modulus:
5 ÷ 3 =
New angle:
𝟐
𝜋
2
∴
𝜋
−3 =
𝑧1
𝑧2
=
H Jackson 2010 / 2015 / 2016 / Academic Skills
5
3
5
3
𝜋
6
𝜋
𝜋
(𝑐𝑜𝑠 (6 ) + 𝑖𝑠𝑖𝑛 (6 ))
4
De Moivres Theorem:
Think: Raise 𝒓 to the
power of 𝑛 and multiply
the angle by 𝑛.
Is used for raising a complex number to a power.
𝒛𝒏 = 𝒓𝒏 (𝒄𝒐𝒔(𝒏𝜽) + 𝒊𝒔𝒊𝒏(𝒏𝜽))
e.g
𝝅
𝝅
If 𝒛 = 𝟑 (𝒄𝒐𝒔 ( 𝟑 ) + 𝒊𝒔𝒊𝒏 ( 𝟑 ))
then 𝑧 5 = 35 (𝑐𝑜𝑠
5𝜋
3
+ 𝑖𝑠𝑖𝑛
5𝜋
3
)
If the power is a fraction (root) there are 𝑛 solutions:
Remember that roots
can be written as
fractions:
Just has a 2𝑘𝜋 extra
𝟏
𝒏
𝟏
𝒏
𝒛 = 𝒓 (𝒄𝒐𝒔 (
𝜽+𝟐𝒌𝝅
𝒏
) + 𝒊𝒔𝒊𝒏 (
𝜽+𝟐𝒌𝝅
𝒏
))
1
𝑛
√𝑧 = 𝑧 𝑛
Where k = 0,1,2…….n-1
If you struggle to remember the theorem for when the power is a fraction – just use the
original (the 1st equation) exactly the same but remember that there will be 𝑛 solutions, all
with the same modulus but with different arguments.
In this case, to find the arguments you need to keep adding
e.g.
𝟐𝝅
𝒏
to your previous answer.
Find the fifth roots of 𝒛 = 𝟐(𝒄𝒐𝒔(𝝅) + 𝒊𝒔𝒊𝒏(𝝅))
Raise 𝒓 to the power of
𝑛 and multiply the
angle by 𝑛.
Using the original equation:
1st answer:
1
1
𝜋
𝜋
𝑧 5 = 25 (𝑐𝑜𝑠 (5 ) + 𝑖𝑠𝑖𝑛 ( 5 ))
(Remember there are 5 answers – so keep adding
2𝜋
5
to find the other 4)
1
1
3𝜋
3𝜋
1
1
5𝜋
5𝜋
1
1
7𝜋
7𝜋
1
1
9𝜋
9𝜋
2nd answer:
𝑧 5 = 25 (𝑐𝑜𝑠 ( 5 ) + 𝑖𝑠𝑖𝑛 ( 5 ))
3rd answer:
𝑧 5 = 25 (𝑐𝑜𝑠 ( 5 ) + 𝑖𝑠𝑖𝑛 ( 5 ))
4th answer:
𝑧 5 = 25 (𝑐𝑜𝑠 ( 5 ) + 𝑖𝑠𝑖𝑛 ( 5 ))
5th answer:
𝑧 5 = 25 (𝑐𝑜𝑠 ( 5 ) + 𝑖𝑠𝑖𝑛 ( 5 ))
H Jackson 2010 / 2015 / 2016 / Academic Skills
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