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Radical solution of Cubic Equation
Cardanos Method
Tariku Getahun
Advisor: Ato Ademe Mekonnen
MiliyonNew Stamp
2 | P a g e
Contents Page
1. inTroduCTion ............................................................................................................................... 3
Complex numbers ...................................................................................................................... 3
Rectangular representation of complex numbers ........................................................................ 3
Operations on complex numbers .................................................................................................. 5
Addition of complex numbers ................................................................................................... 5
Subtraction of complex numbers .............................................................................................. 5
Multiplication of complex numbers .......................................................................................... 5
Complex conjugate and modulus .............................................................................................. 5
Division of complex numbers ..................................................................................................... 6
Polar form of complex numbers ..................................................................................................... 7
Extraction of roots of complex numbers ........................................................................................ 9
2. The radiCal soluTion of CubiC equaTion ...................................................................... 12
Cardanos method ................................................................................................................................... 12
3 | P a g e
inTroduCTion
Complex numbers Definition: A number which is written in the form = + , where , and = 1 is
called a complex number.
i.e. = { + |, , = 1}.
Examples: 1) = 1+ 7 2) = 1 3 3) = 4) = 7
Given a complex number = + , then the real part of is and the imaginary part of is
.
i.e. ()= and ()=
Examples: 1) = 1+ 3, (1+ 3)= 1 and (1+ 3)= 3.
2) = 2 6 , 2 6= 2 and 2 6= 6.
3) = 1+ , (1+ )= 1 and (1+ )= 1.
Rectangular representation of complex numbers
The complex number = + is represented on the plane by an ordered pair = (, ).
4 | P a g e
Example 1: = 2+ = (2,1)
Example 2: = 2 2 = (2, 2)
Example 3: = 3 = (3, 1)
In a complex number we can define only an equal sign = such that there is no order among
complex numbers.
i.e. If we take , , we cannot say < or < .
In fact, the field of complex number is not an ordered field.
5 | P a g e
Definition: Let = + , = + , then we say = if and only if = and
= .
Operations on complex numbers
The operations addition, subtraction, multiplication and division are defined in complex
numbers as follows.
Addition of complex numbers
Let , and = + , = + where , , , .
Then + = + + + = + + ( + ) .
Example: let = 3 2, = 1+
Then + = 3 2 + 1+ = 4 (2 1)= 4 .
Subtraction of complex numbers
Let , and = + , = + where , , , .
Then = + ( + )= + = + ( ) .
Example: let = 5+ 3, = 1
Then = 5+ 3 (1 )= 5 1+ 3 + = 4+ 4.
Multiplication of complex numbers
Let , and = + , = + where , , , .
Then = ( + )( + )= + + +
= + ( + ) .
Example: let = 3 2, = 1+
Then = (3 2)(1+ )= 3+ 3 2 2 = 3+ 3 2 + 2 = 5+ .
But division of complex numbers needs some extra concept is called complex conjugate.
Complex conjugate and modulus
Let = + be a complex number. The conjugate of a complex number denoted by , is
given by = .
Observe that
1. is obtained by reflection of on the real axis.
6 | P a g e
Example: let = 3+ 2, then = 3 2
2. = ( + )( )= + .
The modulus of a complex number = + is denoted by || and given by ||= + .
Therefore ) ||= || , ) = + = || = || , )
= = ()
)
= = (), ) = , ) , , = and + = +
Division of complex numbers
Let , and = + , = + where , , , and 0.
Then
=
=
( + )( )
( + )( )= + +
+
= + + ( )
+
Example: let = 1+ , = 1
Then = 1+ . Thus
=1+
1 1+
1+ =
2
2= 1
7 | P a g e
Polar form of complex numbers
Let = + , , be a complex number with rectangular form. Because of the one-to-
one between set of complex numbers and points in the plane we can represent the complex
number = + in polar form as follow:
= ||= +
= cos
= sin
= + = cos + sin
= (cos + sin)
The representation = (cos + sin) is called polar form of a complex number
= + = (, ), where = tan
. is called an argument of . If is an argument of
then = + 2, is also an argument of .
Propositon: If = (cos + sin), = (cos + sin), then
i. = [cos( + )+ sin( + )]
ii.
=
[cos( )+ sin( )]
Proof: i) = [(cos + sin)][(cos + sin)]
= cos cos + cos sin + sin cos sin sin
= [cos cos sin sin + (cos sin + sin cos)]
= [cos( + )+ sin( + )]
ii)
=
=
=cos cos sin cos + sin cos + sin sin
cos + sin
=cos cos + sin sin + (sin cos sin cos)
1
=[cos( )+ sin( )]
8 | P a g e
Theorem: (Demoivres Formula)
If = [cos + sin], then = [cos()+ sin()].
Proof: By induction
i) For = 2, = = [(cos()+ sin())][(cos()+ sin())]
= cos + cos sin + sin cos sin
= [cos sin + 2 cos sin]= [cos2 + sin2]
ii) Suppose it is true for = , i.e. = (cos()+ sin())
iii) WTS it is true for = + 1, i.e. = cos( + 1)+ sin( + 1)
Now, = = [cos + sin][cos()+ sin()]
= [cos + sin][cos()+ sin()]
= [cos cos()+ sin cos()+ cos sin() sin sin()]
= [cos cos() sin sin()+ (sin cos()+ cos sin())]
= [cos cos() sin sin()+ (sin cos()+ cos sin())]
= [cos( + 1) + sin( + 1)]
= [cos( + 1) + sin( + 1)]
= [cos( + 1) + sin( + 1)]
By principle of mathematical induction
= [cos + sin],
Example 1: = 3 , find =?
Solution:||= 3+ (1) = 2, = tan
=
.
Let = (cos()+ sin()), where ||=
Then = 2cos
+ sin
Therefore = 2 cos
+ sin
= 2 cos
+ sin
= 2
+
9 | P a g e
Example 2: = 2+ 2, find =?
Solution:||= 8, = tan(2/2)=
.
Let = (cos()+ sin()), where ||=
Then = 8cos
+ sin
Therefore = 8cos4
+ sin4
= 64(cos()+ sin())= 64.
Extraction of roots of complex numbers
Suppose and are complex numbers, then is the root of iff = , . In this
case =
.
Let = (cos + sin), = || and = (cos + sin), = | | such that = .
Then = [cos + sin]= [cos + sin]
= and = + 2, .
= and =
, .
If = , then = cos
+ sin
, .
But are distinct if we restrict = 0,1,2, , 1.
Example 1: Find third roots of unity such that = 1.
Solution: Let = 1, | |= 1, = 0, 1 = cos0+ sin0
Let = (cos + sin), then = [cos3 + sin3]= cos0+ sin0
= 1 and 3 = 0+ 2, = 0,1,2
= 1 and =
,
= cos
+ sin
, = 0,1,2
= 1, =
+
, =
10 | P a g e
Example 2: Find fourth roots of such that = .
Solution:
Let = , | |= 1, = /2, = cos
+ sin
Let = (cos + sin), then = [cos4 + sin4]= cos
+ sin
= 1 and 4 =
+ 2, = 0,1,2,3
= 1 and =
, = 0,1,2,3
= cos
2
+ sin
2
= 0,1,2,3
Then we have , , , distinct roots.
= cos
8+ sin
8= 0.923+ 0.382, = cos
5
8+ sin
5
8= 0.382+ 0.923
= cos9
8+ sin
9
8= 0.923 0.382, = cos
13
8+ sin
13
8= 0.382 0.923
Example 3: Find all 6 roots of 1 such that = 1.
Solution:
Let = 1, | |= 1 = , = , 1 = cos + sin
Let = (cos + sin), then = [cos6 + sin6]= cos + sin
= 1 and 6 = + 2, = 0,1,2,3,4,5
= 1 and =
, = 0,1,2,3,4,5
= cos
+ sin
, = 0,1,2,3,4,5
Then we have , , , , , distinct roots.
=3
2+
2, = , =
3
2+
2, =
3
2
2
= 1, =3
2
2
11 | P a g e
Example 4: Find all 4 roots of 3 33 such that = 3 33 .
Solution:
Let = 3 33 , | |= 6 = , =
, 3 33 = cos
+ sin
Let = (cos + sin), then = [cos4 + sin4]= 6
cos
+ sin
= 6 and 4 =
+ 2, = 0,1,2,3
= 6
and =
, = 0,1,2,3
= 64
cos4
3
+ sin
4
3
, = 0,1,2,3
Then we have , , , distinct roots.
= 64
cos
3+ sin
3 = 6
41
2+3
2 , = 6
4cos
5
6+ sin
5
6 = 6
4
3
2+
2,
= 64
cos4
3+ sin
4
3 = 6
43 33 , = 6
4(cos2 + sin2)= 6
4
12 | P a g e
The radiCal soluTion of CubiC equaTion
Cardanos method
The general cubic equation is given by
+ + + = 0 (1)
Dividing (1) by gives us
+
+
+
= 0 (2)
Now, substitute =
3
+
3
+
3 +
= 0
+
3
27 +
2
3 +
9 +
3 +
= 0
+
3 2
3+
+
9
27
3+
= 0
+ 3
3 +
2 9+ 27
27 = 0
Then we get the transformed equation
+ + = 0 (3)
where =
and
.
We call the equation in (3) depressed equation.
Consider 3 + + = 0.
Introduce two variables and linked by the candidate + = and substitute in (3).
Then we get
( + ) + ( + )+ = 0
+ + 3 + 3 + + + = 0
+ + 3 ( + )+ ( + )+ = 0
+ + (3 + )( + )+ = 0 ()
13 | P a g e
We now choose and so that 3 + = 0
=
3, =
3
Since 3 + = 0, () becomes
+ + = 0 ()
By substituting =
, we get
27+ = 0
Then multiplying by both side gives
+
27= 0
This a quadratic equation with such that
() +
27= 0
By using general quadratic formula
=
2
2
+
3
The expression =
+
is the discriminant.
Then
=
2
Since + + = 0 = = (
)
=
2
This shows and are algebraically conjugate expressions.
Case 1: If > 0 i.e. +
> 0
Now =
+ =
14 | P a g e
Let =
+
= and =
= , where and are reals.
= =
By using De moivres formula we can solve for as follow
= (cos3 + sin3)= (cos0+ sin0)
= = and =
, .
= cos2
3 + sin
2
3 ,
We have three distinct roots
= = , = cos2
3 + sin
2
3 =
1
2+3
2
= cos4
3 + sin
4
3 =
1
23
2
And
Since and are algebraically conjugate of each other.
= (cos3 sin3)= (cos0 sin0)
= = and =
, .
= cos2
3 sin
2
3 ,
We have three distinct roots
= = , = cos2
3 sin
2
3 =
1
23
2
= cos4
3 sin
4
3 =
1
2+3
2
If , and are roots of + + = 0, then
= +
To get and we have to show =
.
For , =
+
=
+
+
=
15 | P a g e
But =
+
and =
So we have
= =
2+
2+
=
2+
2
=
4
=
4
2
+
3
=
3
=
3
=
. Hence = + =
( + )+
( )
For , =
+
=
+
+
=
= =
3
Hence = + =
( + )
( ).
Therefore
= + = + , = 1
2( + )+
3
2( ), =
1
2( + )
3
2( ).
where =
+
, =
and =
+
.
It follows the roots of + + + = 0 are given as
=
3, = 1,2,3
Where are distinct roots of + + = 0, where =
and
.
Case 2: If = 0, = + becomes =
Since + + = 0 = =
=
=
= = /2
= = , then = cos
+ sin
16 | P a g e
= = /2
= 1
2+3
2 = /2
1
2+3
2
= 1
23
2 = /2
1
23
2
In a similar way we can solve for
= cos2
3 sin
2
3
= = /2
= 1
23
2 = /2
1
23
2
= 1
2+3
2 = /2
1
2+3
2
= + = + = + = 2 = 2/2
= + = /2
= + = /2
= 2/2 , = = /2
are roots of + + = 0, hence = =
; = 1,2,3
is the solution of the general cubic equation.
Case 3: If < 0, if +
< 0 we have three real roots given by =
Example 1: Solve for
+ 6 20 = 0
Solution: This equation is of the form + + = 0
Then = 6, = 20
=
2+
2
+
3
== 20
2 +
20
2
+ 6
3
= 10+ 63
17 | P a g e
=
2
2
+
3
== 20
2
20
2
+ 6
3
= 10 63
= 10+ 63
, = 10 63
Let = , =
Then as we have solved for > 0 in case 1 above
= + = 10+ 63
+ 10 63
= 1
2( + )+
3
2( )=
1
210+ 63
+ 10 63
+3
2 10+ 63
10 63
= 1
2( + )
3
2( )=
1
210+ 63
+ 10 63
3
2 10+ 63
10 63
Example 1: Solve for
7 + 6 = 0
Solution: This equation is of the form + + = 0 with = 7 = 6, < 0
= 3+ 100
27,
= 3 100
27
The three roots of 7 + 6 = 0 are = 1 = 3 = 2 which are real.