Radical solution of Cubic Equation

Cardanos Method

Tariku Getahun

Advisor: Ato Ademe Mekonnen

MiliyonNew Stamp

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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

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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 = (, ).

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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.

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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.

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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

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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( )]

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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

+

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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, =

+

, =

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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

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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

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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 ()

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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 =

+ =

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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 , =

+

=

+

+

=

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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

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= = /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

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=

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.