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Transcript
Complex Numbers Summary
Academic Skills Advice
What does a complex number mean?
A complex number has a β€˜real’ part and an β€˜imaginary’ part (the imaginary part involves the
square root of a negative number).
e.g. 𝒁 = 𝒙 + π’Šπ’š
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 βˆšβˆ’1
so:
2
𝑖 2 = (βˆšβˆ’1) = -1
𝑖 4 = (𝑖 2)2 = (-1)2 = 1
Summary:
π’Š = βˆšβˆ’πŸ
π’ŠπŸ = βˆ’πŸ
π’ŠπŸ’ = 𝟏
For any power of 𝑖 substitute as many 𝑖 2’s as possible
and they will all end up as ±π‘– or ±1.
Examples:
𝑖 20 = (𝑖 2 )10 = (βˆ’1)10 = 1
𝑖 33 = (𝑖 2 )16 𝑖 = (βˆ’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 = βˆ’1).
Example:
(3 + 5𝑖 )(4 – 2𝑖 )
2
= 12 – 6𝑖 + 20𝑖 – 10𝑖
= 12 + 14𝑖 – 10 (βˆ’1)
= 22 + 14𝑖
H Jackson 2010 / 2015 / 2016 / Academic Skills
1
Except where otherwise noted, this work is licensed under
http://creativecommons.org/licenses/by-nc-sa/3.0/
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(βˆ’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
4βˆ’5𝑖
3+2𝑖
=
3
=
4βˆ’5𝑖
2
4
+ 𝑖 = 1.5 + 2𝑖
2
3+2𝑖
×
3βˆ’2𝑖
3βˆ’2𝑖
=
12βˆ’8π‘–βˆ’15𝑖+10𝑖 2
9βˆ’6𝑖+6π‘–βˆ’4𝑖 2
=
12βˆ’23𝑖+10(βˆ’1)
9βˆ’4(βˆ’1)
=
2βˆ’23𝑖
13
=
𝟐
πŸπŸ‘
βˆ’
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.
* When finding πœƒ either:
Imaginary
P (𝑧
= π‘₯ + 𝑦𝑖)
y
NB tan(ΞΈ) =
𝑦
so
π‘₯
π’š
𝜽 = π’•π’‚π’βˆ’πŸ ( )
𝒙
0
πœƒ
π‘₯
ο‚·
ο‚·
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 π‘Ÿ and is the length of OP, therefore |𝒛| = 𝒓 = βˆšπ’™πŸ + π’šπŸ
Argument: is the angle θ that is made with the horizontal axis (denoted by ∠).
Example:
P (𝑧
= 4 + 5𝑖)
|𝒛| = βˆšπŸ’πŸ + πŸ“πŸ = βˆšπŸ’πŸ = πŸ”. πŸ’
πŸ“
5
𝜽 = π’•π’‚π’βˆ’πŸ ( ) = 𝟎. πŸ— radians
πŸ’
0
πœƒ
4
H Jackson 2010 / 2015 / 2016 / Academic Skills
Real
2
Polar & Exponential Form
As well as the cartesian form (𝑧 = π‘₯ + 𝑖𝑦) there are 2 alternative ways of writing a complex
number:
𝒛 = 𝒓(π’„π’π’”πœ½ + π’Šπ’”π’Šπ’πœ½)
Polar:
Where:
π‘Ÿ is the modulus (length of the line) |𝑧|
πœƒ is the argument (angle it makes with the π‘₯-axis)
Exponential: 𝒛 = π’“π’†π’Šπœ½
(this should be in radians for the exponential form).
Remember:
To find the modulus (length), 𝒓:
use Pythagoras
π’š
To find the argument (angle), 𝜽: use π’•π’‚π’βˆ’πŸ (𝒙) *
Converting between the different forms:
Cartesian
Need to find 𝒓 and 𝜽
π‘Ÿ = √π‘₯ 2 + 𝑦 2
Polar or Exponential
𝑦
πœƒ = π‘‘π‘Žπ‘›βˆ’1 ( )
π‘₯
Polar or Exponential
Example:
0
Need to find 𝒙 and π’š
π‘₯ = π‘Ÿπ‘π‘œπ‘ πœƒ
𝑦 = π‘Ÿπ‘ π‘–π‘›πœƒ
Cartesian
Express 𝒛 = πŸ‘ + πŸ’π’Š in polar and exponential form
imaginary
π‘Ÿ
*
4
πœƒ
3
Modulus:
π‘Ÿ = √32 + 42 = √16 + 9 = √25 = 5
Argument:
πœƒ = π‘‘π‘Žπ‘›βˆ’1 (3) = 53.1o
Polar form:
𝑧 = 5(cos(53.1) + 𝑖𝑠𝑖𝑛(53.1))
Exp form:
𝑧 = 5𝑒 0.927𝑖
real
4
(in radians πœƒ = 0.927π‘Ÿ )
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
= πŸ•π’†π’ŠπŸ‘
in cartesian form
πœ‹
π‘₯ = π‘Ÿπ‘π‘œπ‘ πœƒ
∴ π‘₯ = 7 cos (3 ) = 3.5
𝑦 = π‘Ÿπ‘ π‘–π‘›πœƒ
∴ 𝑦 = 7 sin ( 3 ) = 6.1
Cartesian form:
πœ‹
𝑧 = 3.5 + 6.1𝑖
H Jackson 2010 / 2015 / 2016 / Academic Skills
3
A reminder of the 3 forms:
Cartesian
𝒛 = 𝒙 + π’Šπ’š
Polar
𝒛 = 𝒓(π’„π’π’”πœ½ + π’Šπ’”π’Šπ’πœ½)
π‘Ÿ = √π‘₯ 2 + 𝑦 2
π‘₯ = π‘Ÿπ‘π‘œπ‘ πœƒ
𝑦 = π‘Ÿπ‘ π‘–π‘›πœƒ
Conversions:
Exponential
𝒛 = π’“π’†π’Šπœ½
𝑦
πœƒ = π‘‘π‘Žπ‘›βˆ’1 ( ) *
π‘₯
Multiplying with Polar or Exponential form
Let
This means, when multiplying 2 complex numbers:
𝑧1 = 𝑧2 𝑧3
Multiply the π‘Ÿβ€²π‘ 
Add the angles (πœƒβ€²π‘ )
Then |𝑧1 | = |𝑧2 | × |𝑧3 |
And βˆ π‘§1 = βˆ π‘§2 + βˆ π‘§3
𝝅
𝝅
Example:
If π’›πŸ = πŸ“π’†πŸ π’Š and π’›πŸ = πŸ‘π’†πŸ‘ π’Š find π’›πŸ π’›πŸ
New modulus:
New angle:
5 × 3 = 15
πœ‹
πœ‹
5πœ‹
+
=
2
3
6
5πœ‹
∴ 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 π‘Ÿβ€²π‘ 
Subtract the angles (πœƒβ€²π‘ )
|𝑧2 |
|𝑧3 |
βˆ π‘§1 = βˆ π‘§2 βˆ’ βˆ π‘§3
𝝅
𝝅
𝝅
𝝅
𝒛
Example:
if π’›πŸ = πŸ“ (𝒄𝒐𝒔 (𝟐 ) + π’Šπ’”π’Šπ’ ( 𝟐 )) and π’›πŸ = πŸ‘ (𝒄𝒐𝒔 ( πŸ‘ ) + π’Šπ’”π’Šπ’ ( πŸ‘ )) find π’›πŸ
New modulus:
5 ÷ 3 =
New angle:
𝟐
πœ‹
2
∴
πœ‹
βˆ’3 =
𝑧1
𝑧2
=
H Jackson 2010 / 2015 / 2016 / Academic Skills
5
3
5
3
πœ‹
6
πœ‹
πœ‹
(π‘π‘œπ‘  (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(π’πœ½) + π’Šπ’”π’Šπ’(π’πœ½))
e.g
𝝅
𝝅
If 𝒛 = πŸ‘ (𝒄𝒐𝒔 ( πŸ‘ ) + π’Šπ’”π’Šπ’ ( πŸ‘ ))
then 𝑧 5 = 35 (π‘π‘œπ‘ 
5πœ‹
3
+ 𝑖𝑠𝑖𝑛
5πœ‹
3
)
If the power is a fraction (root) there are 𝑛 solutions:
Remember that roots
can be written as
fractions:
Just has a 2π‘˜πœ‹ extra
𝟏
𝒏
𝟏
𝒏
𝒛 = 𝒓 (𝒄𝒐𝒔 (
𝜽+πŸπ’Œπ…
𝒏
) + π’Šπ’”π’Šπ’ (
𝜽+πŸπ’Œπ…
𝒏
))
1
𝑛
βˆšπ‘§ = 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.
πŸπ…
𝒏
to your previous answer.
Find the fifth roots of 𝒛 = 𝟐(𝒄𝒐𝒔(𝝅) + π’Šπ’”π’Šπ’(𝝅))
Raise 𝒓 to the power of
𝑛 and multiply the
angle by 𝑛.
Using the original equation:
1st answer:
1
1
πœ‹
πœ‹
𝑧 5 = 25 (π‘π‘œπ‘  (5 ) + 𝑖𝑠𝑖𝑛 ( 5 ))
(Remember there are 5 answers – so keep adding
2πœ‹
5
to find the other 4)
1
1
3πœ‹
3πœ‹
1
1
5πœ‹
5πœ‹
1
1
7πœ‹
7πœ‹
1
1
9πœ‹
9πœ‹
2nd answer:
𝑧 5 = 25 (π‘π‘œπ‘  ( 5 ) + 𝑖𝑠𝑖𝑛 ( 5 ))
3rd answer:
𝑧 5 = 25 (π‘π‘œπ‘  ( 5 ) + 𝑖𝑠𝑖𝑛 ( 5 ))
4th answer:
𝑧 5 = 25 (π‘π‘œπ‘  ( 5 ) + 𝑖𝑠𝑖𝑛 ( 5 ))
5th answer:
𝑧 5 = 25 (π‘π‘œπ‘  ( 5 ) + 𝑖𝑠𝑖𝑛 ( 5 ))
H Jackson 2010 / 2015 / 2016 / Academic Skills
5