Development Teachers’ Mathematics Competency through Teaching Complex Numbers in High School in Vietnam

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h

National Academy of Education Management, Hanoi city, Vietnam

Corresponding Author: Tran Trung Tinh, [email protected]

ABSTRACT: In teaching mathematics at high schools, the mathematics competency of teacher is an important procedure in teaching process. This paper, I pointed out some application of complex numbers in mathematics. And i hope it will help the teachers to develop high school their mathematics competency in Vietnam.

KEYWORDS: Teachers’ mathematics competency,

complex numbers, teaching methods.

1. BUILDING A

C

FIELD

We review Carte

T

=

R



R

=

{(

a

,

b

)

|

a

,

b



R

}

and the following definition:

d

b

c

a

d

c

b

a

,

)

=

(

,

)

if

and

only

if

=

,

=

(

)

,

(

=

)

,

(

)

,

(

a

b



c

d

a



c

b



d

).

,

(

=

)

,

).(

,

(

a

b

c

d

ac



bd

ad



bc

We can write,

(

a

,

b

).(

c

,

d

)

through

(

a

,

b

)(

c

,

d

).

From the definition of multiplication:

• Where

i

=

(0,1)



T

we have

1,0)

(

=

(0,1)(0,1)

=

.

=

2



i

i

i

•

(

a

,

b

)(1,0)

=

(

a

,

b

)

=

(1,0)(

a

,

b

)

• (a,b)=(a,0)(0,b)=(a,0)(b,0)(0,1),(a,b)T.

Notation

C

is set T with Mathematical operations above. We get:

Lemma 1. Map



:

R



C

,

a



(

a

,0),

Is a single map and such that

)

(

)

(

=

)

(

),

(

)

(

=

)

(

a



a





a





a





a

a





a



a





forall

.

,

a

R

a





C

a

,0)



(

is the same as

a



R

.

Then we have

bi

a

b

a

b

a

,

)

=

(

,0)



(

,0)(0,1)

=



(

where

1.

=

1,0)

(

=

2





i

Therefore i or a or

a



bi

are equal in

C

.

Deduce

C

=

{

a



bi

|

a

,

b



R

,

i

2

=



1}

and in

C

we have the following results:

d

b

c

a

di

c

bi

a



=



if

and

only

if

=

,

=

i

d

b

c

a

di

c

bi

a







=





(



)

.

)

(

=

)

)(

(

a



bi

c



di

ac



bd



ad



bc

i

This objects

z

=

a



bi



C

considered as a

complex number with a is the real part (Notation

),

(

z

Re

) and

b

is the imaginary part, (Notation

)

(

z

Im

); and i is the imaginary unit. Complex number

a



bi

is called a conjugate complex of

bi

a

z

=



and notation

z

=

a



bi

.

We have

2 1 2 1 2 2

. = , =

) )( (

= a bi a bi a b z z z z z

z    and

The modulus of a complex number

z

is

z

z

z

|=

|

. The argument number of a complex number

z



=

c



di

is



z



=



c



di

, then we have

.

)

(

=

)

(

)

(

=

a

bi

c

di

a

c

b

d

i

z

z

















Consider the system of coordinate

(

Oxy

).

This Complex number

z

=

a



bi

we get a point

).

;

(

a

b

M

and it is a bijection

R

R

C





).

;

(

=

a

bi

M

a

b

z





In mathematics, the complex plane or z-plane is a geometric representation of the complex numbers established by the real axis and the perpendicular imaginary axis. with the real part of a complex number represented by a displacement along the x-axis, and the imaginary part by a displacement along the y-axis.

In geometry, a seventeen-sided polygon is a constructible polygon (that is, one that can be constructed using a compass and unmarked straightedge). This was shown by Carl Friedrich Gauss in 1796 at the age of 19.

Proposition 1. R field is in

C

field

Proof: We see that

C

is a commutative ring, where 1 is unit. Suppose

z

=

a



bi



0.

then

0.

>

2 2

b

a



Suppose

z



=

x



yi



C

must satisfy requirements zz=1 or











0.

=

1

=

ay

bx

by

ax

We get

.

=

,

=

2 2 2

2

_a

_b

b

y

b

a

a

x







Deduce

i

b

a

b

b

a

a

z

=

_{2 2 2 2}









is a inverse number of z.

with

a



0

i



C

and we can say that

R



C

.

Notation that, inverse number of

z



0

is

2 1

|

|

=

z

z

z

 và

.

|

|

=

=

2 1

z

z

z

z

z

z

z



_





Definition 1. [N+15] Given complex number

z



0.

Suppose M is a point in the complex plane for expression z. The angle measurement created by first ray

Ox

and last ray

OM

called Argument of

z

and notation

Arg

(

z

).

The angle

,

,

=









xOM









called argument of

z

and notation

Argz

.

Argument of 0 complex is not definition.

Notation that, if



is a argument of

z

then forall argument of

z

are





k

.2



where

k



Z

.

If

0,



z

noindent





k

.2



is Argument of z.

Notation r= zz. Therefore, the complex number

bi

a

z

=



has

a

=

r

cos



,

b

=

r

sin



.

Deduce, if

0



z

then

z

=

r

(

cos





i

sin



)

, we called

dạng lượng giác of z.

Proposition 2. Given complex number

z

₁

,

z

₂ and

),

sin

cos

(

=

_{1 1 1}

1

r



i



z



),

sin

cos

(

=

_{2 2 2}

2

r



i



z



r

₁

,

r

₂



0,

we always have

•

|

z

₁

z

₂

|=|

z

₁

||

z

₂

|,

|

|

|

|

|=

|

2 1

2 1

z

z

z

z

và

.

|=

|

|

|

|

|

z

₁



z

₂



z

₁



z

₂

r

₁



r

₂

•

z

1

z

2

=

r

1

r

2

[

cos

(



1





2

)



i

sin

(



1





2

)]

• = [cos( 1 2) sin( 1 2)]

2 1

2

1















i r

r z z

where

0.

>

2

r

Proof: It’s easier to get the result.

Example 1. With

a



bi

=

(

x



iy

)

n we have

.

)

(

=

2 2

2

2 n

y

x

b

a





Proof: From

a



bi

=

(

x



iy

)

n deduce

.

)

(

=

x

iy

n

bi

a





Therefore

.

)

(

=

2 2

2

2 n

y

x

b

a





Proposition 3. [Moivre] If

z

=

r

(

cos





i

sin



)

then forall

n



 we have

)].

(

sin

)

(

cos

[

=

r

n



i

n



z

n n



Proof: Using Mathematical induction methods by n to get the result.

Corollary 1. An nth root of a complex number

0

)

sin

cos

(

=

r





i





z

are different values in

)

2

sin

2

cos

(

=

1/

n

k

i

n

k

r

z

_k n















where

.

,

1,2,

=

n

k



Next, we prove a Euler’s famous theorem below:

Theorem 1. [Euler] For all x we have

.

sin

cos

=

x

i

x

e

ix



Proof: From

















!

)

(

3!

)

(

2!

)

(

1!

1

=

3 2

n

ix

ix

ix

ix

e

n ix

we

deduce

)

)!

(2

1)

(

4!

2!

(1

=

2 4

2

















n

x

x

x

e

n n ix

).

1)!

(2

1)

(

5!

3!

(

1 2 1 5

3





















 

n

x

x

x

x

i

n n

Therefore

e

ix

=

cos

x



i

sin

x

.

Corollary 2. For all Real numbers x,y we always have

• ( )

=

ix y

iy ix

e

e

e

 and

(

e

ix

)

n

=

e

inx forall

n



Z

.

• ( )

=

i x y

iy ix

e

e

e



and ix

=

ix

=

1

_ix

.

e

e

e



•

.

2

=

sin

,

2

=

cos

i

e

e

x

e

e

x

ix ix ix

ix









•

(

)

=

cos

(

).

2

1

>=

,

<

x

y

e

e

e

e

e

e

_ix

iy

iy ix iy

ix





Proof: From theorem 1, we easier to get the result. Given three distinctive points

A

,

B

,

C

corresponding to three complex numbers

a

,

b

,

c

.

We notation

C

b

c

a

c

C

B

A







=

]

,

,

[

and we call it

single ratio of point sets

A

,

B

,

C

.

We have result.

Corollary 3. Given three distinctive points

A

,

B

,

C

corresponding to three complex numbers

a

,

b

,

c

all on a line if and only if

[

,

,

]

=

R

.

b

c

a

c

C

B

A







Proof: Set

c



a

=

re

i

,

c



b

=

r



e

i

.

then we have

.

=

=

]

,

,

[

()







i

e

r

r

b

c

a

c

C

B

A

three points

A

,

B

,

C

all on a line if and only if

k



b

c

a

c

=

arg





or

.

=

r

r

e

r

r

ik







 Therefore, three distinctive points

C

B

A

,

,

corresponding to three complex numbers

c

b

a

,

,

all on a line if and only if

.

=

]

,

,

[

R

b

c

a

c

C

B

A







).

(

2

1

=

]

,

[

),

(

2

1

>=

,

<

_{1 2 1 2 1 2 1 2}

z

₁

z

₂

z

₁

z

₂

i

z

z

z

z

z

z

z

z





Proposition 4. If

)

sin

cos

(

=

),

sin

cos

(

=

1 1 1 2 2 2 2

1

r



i



z

r



i



z





where

r

₁

,

r

₂



0

then

•

|

z

₁

z

₂

|=|

z

₁

||

z

₂

|,

| |

| | |= |

2 1

2 1

z z z z

and

.

|

|

|

|

|

|

z

1



z

2



z

1



z

2

•

z

₁

z

₂

=

r

₁

r

₂

[

cos

(



₁





₂

)



i

sin

(



₁





₂

)]

• = [cos( _{1 2}) sin( _{1 2})]

2 1

2

1







_i







r r z z

where

0.

>

2

r

•

<

z

₁

,

z

₂

>=|

z

₁

||

z

₂

|

cos

(



₁





₂

)

and

.

>

,

>=<

,

<

z

1

z

2

z

2

z

1

•

<

az

₁



bz

₃

,

z

₂

>=

a

<

z

₁

,

z

₂

>



b

<

z

₃

,

z

₂

>

forall complex numbers

z

₁

,

z

₂

,

z

₃ and forall

.

,

b

R

a



•

[

z

1

,

z

2

]

=|

z

1

||

z

2

|

sin

(



2





1

)

and

].

,

[

=

]

,

[

z

₁

z

₂



z

₂

z

₁ • Given

2 2

2 1 1

1

=

cos



i

sin



,

z

=

cos



i

sin



z





we

have

)

2

sin

2

cos

(

2

sin

2

=

1 2 1 2 1 2

2 1























z

i

i

z

.

|

2

sin

|

2

|=

|

1 2

2 1









z

z

Proof: It’s easier to get the result.

You can check all the results. Special, deviation product hasgeometric significance: Given point

O

and the coordinates of the points

M

,

N

are respectively

z

1

,

z

2

,

three points

O

,

M

,

N

all on a line if and only if

[

z

₁

,

z

₂

]

=

0.

When three points

N

M

O

,

,

not in line then absolute value of deviation product

[

z

₁

,

z

₂

]

twice the area triangle

.

OMN

Given two complex numbers

z

₁ and

z

₂ we always have

.

,

2

arg

=

arg

|,

|=|

|

=

_{2 1 2 1 2}

1

z

z

z

z

z

k

k

Z

z









.

,

2

)

(

arg

)

(

arg

=

)

(

arg

z

₁

z

₂

z

₁



z

₂



k



k



Z

.

,

2

)

(

arg

)

(

arg

=

)

(

arg

_{1 2}

2 1

Z

k

k

z

z

z

z

_

_

_

_

).

(

)

(

=

)

(

z

₁

z

₂

Arg

z

₁

Arg

z

₂

Arg



).

(

)

(

=

)

(

_{1 2}

2

1

_Arg

_z

_Arg

_z

z

z

Arg



We say that, the polynomials of positive degree in

]

[

x

C

always have roots in

C

.

The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. This includes polynomials with real coefficients, since every real number is a complex number with an imaginary part equal to zero.

Equivalently (by definition), the theorem states that the field of complex numbers is algebraically closed.

Definition 2. In abstract algebra, an algebraically closed field K contains a root for every non-constant polynomial in

K

[

x

]

, the ring of polynomials in the variable x with coefficients in

K.

From The fundamental theorem of algebra, we deduce results of irreducible polynomial in

C

[

x

]

:

Corollary 4. Given the polynomials in

C

[

x

]

where degree

n

>

0

have n roots in

C

and the irreducible polynomials in

C

[

x

]

have 1 degree.

Corollary 5. Let

f

(

x

)



R

[

x

]

\

R

.

f

(

x

)

is irreducible polynomial if and only if or

b

ax

x

f

(

)

=



where

a



0

or

c

bx

ax

x

f

(

)

=

2





where

a



0

and

0.

<

4

2

ac

b



2. SOME EXAMPLES OF COMPLEX

NUMBERS

Example 2. Prove that, with two complex numbers

1

z

and

z

₂ we always have

.

|

|

|

=|

|

|

2

|

|

2

z

₁ 2



z

₂ 2

z

₁



z

₂ 2



z

₁



z

₂ 2

Proof: Assume

A

(

z

1

),

B

(

z

2

)

and

C

(

z

1



z

2

).

So

the quadrangle

OACB

is Parallelogram, we deduce

2 2

2 2

2

2

=

OA

OB

AB

OC





or

.

|

|

|

=|

]

|

|

|

2[|

z

₁ 2



z

₂ 2

z

₁



z

₂ 2



z

₁



z

₂ 2

Example 3. With two complex numbers

z

and z, we set

u

=

z

z



.

Prove that

.

|

2

|

|

2

|=|

|

|

|

z



z



z



z





u



z



z





u

Proof: Because

| 2

) (

| | 2

) (

|=| 2

| | 2 |

2 2

z z z

z u z z u z

z _{ }  _   _  

if we set z1= z,z2 = z then we will have to prove:

2[|

z

₁

|

2



|

z

₂

|

2

]

=|

z

₁



z

₂

|

2



|

z

₁



z

₂

|

2

.

We consider

A

(

z

1

),

B

(

z

2

)

and

C

(

z

1



z

2

).

Because the quadrangle

OACB

is Parallelogram,

deduce

OC

2



AB

2

=

2

OA

2



2

OB

2 or

.

|

|

|

=|

]

|

|

|

Example 4. [N+15]Prove that, with three different complex numbers

z

₁

,

z

₂

,

z

₃ have

|

2

|

2

|

2

|

2

1 3 _{1 3}

2 1 2

1

z

_z

_z

z

z

_z

_z

z



_

_



_

.

|

2

4

|

2

|

2

=|

2 3 _{1 2 1 3} 2 3

1 3 2 3

2

z

_z

_z

_z

z

z

_z

_z

_z

_z

z

z

z

















Proof: Set u_j = z_j where

j

=

1,2,3.

We will have to prove

.

|

2

|

2

2

|

|

=

|

|

|

|

2 3 2

1 2 3 2 2 3 1 2 2 1

u

u

u

u

u

u

u

u

u















Consider triangle

ABC

where

)

(

),

(

),

(

u

₁

B

u

₂

C

u

₃

A

and

)

2

(

u

2

u

3

M



is midpoint

of sided

BC

.

Because

2

2

=

2 2 2 2

a

m

c

b



_a



deduce we get

2 3 2 1 2 3 2 2 3 1 2 2 1

|

2

|

2

2

|

|

=

|

|

|

|

u



u



u



u

u



u



u



u



u

and hence, ends the proof.

Example 5. [Euler] Given

R

y

y

y

y

x

x

x

x

₁

,

₂

,

₃

,

₄

,

₁

,

₂

,

₃

,

₄



we always have

, =

) )(

(x₁2x₂2x₃2x₄2 y₁2y₂2y₃2y₄2 u₁2u₂2u₃2u₄2 where







































1 4 2 3 3 2 4 1 4 2 4 1 3 4 2 3 1 3 3 4 4 3 1 2 2 1 2 4 4 3 3 2 2 1 1 1

=

=

=

=

y

x

y

x

y

x

y

x

u

y

x

y

x

y

x

y

x

u

y

x

y

x

y

x

y

x

u

y

x

y

x

y

x

y

x

u

and deduce inequality

. ) ( _{1 1 2 2 3 3 4 4} 2 2 4 2 3 2 2 2

1 u u u x y x y x y x y

u       

Proof: Set

z

₁

=

x

₁



ix

₂

,

z

₂

=

x

₃



ix

₄

,

z

₃

=

y

₁



iy

₂ and

z

4

=

y

3



iy

4

.

We have

) )(

(

= x₁2 x₂2 x₃2 x₄2 y₁2 y₂2 y₃2 y₄2

T      

. =

) )(

(

= z₁z₁z₂z₂ z₃z₃z₄z₄ z₁z₃z₁z₃z₂z₃z₂z₃z₁z₄z₁z₄z₂z₄z₂z₄ Therefore T =u₁2u₂2 u₃2u₄2.

Example 6. Given three different complex numbers

3 2 1

,

z

,

z

z

we have

n n n n

x

z

z

z

z

z

x

z

x

z

z

z

z

z

z

x

z

x

z

z

z

z

z

z

x

z

x

z

x

f

=

)

)(

(

)

)(

(

)

)(

(

)

)(

(

)

)(

(

)

)(

(

=

)

(

2 3 1 3 2 1 3 1 2 3 2 1 3 2 3 1 2 1 3 2 1





























với

n

=

0,1,2.

Proof: The polynomial

f

(

x

)



x

n and its degree is less than

3

such that

f

(

z

₁

)

=

f

(

z

₂

)

=

f

(

z

₃

)

=

0.

Therefore

f

(

x

)

=

x

n and we get the result.

Example 7. Given three different complex numbers

3 2 1

,

z

,

z

z

and two complex numbers u,vwe have

)

)(

)(

(

)

)(

(

=

)

)(

)(

(

)

)(

(

1 3 1 2 1 1 1 3 2

1

z

z

z

z

x

z

v

z

u

z

z

x

z

x

z

x

v

x

u

x





















. ) )( )( ( ) )( ( ) )( )( ( ) )( ( 3 2 3 1 3 3 3 2 3 2 1 2 2 2 z x z z z z v z u z z x z z z z v z u z            

Proof: We consider

. = ) )( )( ( ) )( ( 3 3 2 2 1 1 3 2

1 x z

x z x x z x x z x z x z x v x u x          

Then we have

).

)(

(

)

)(

(

)

)(

(

=

)

)(

(

x



u

x



v

x

₁

x



z

₂

x



z

₃



x

₂

x



z

₃

x



z

₁



x

₃

x



z

₁

x



z

₂ Where

x

=

z

₁

,

z

₂

,

z

₃ we get

                     ) )( ( ) )( ( = ) )( ( ) )( ( = ) )( ( ) )( ( = 2 3 1 3 3 3 3 1 2 3 2 2 2 2 3 1 2 1 1 1 1 z z z z v z u z x z z z z v z u z x z z z z v z u z x

and we get the result.

Example 8. Given three different complex numbers

3 2 1

,

z

,

z

z

we have

) )( )( ( ) )( )(

( _{2 3 2 1 2}

1 3 1 3 1 2 1 3 2 t z z z z z z z t z z z z z z z         

.

)

)(

)(

(

=

)

)(

)(

(

_{1 2 3}

3 2 1 3 2 3 1 3 2 1

z

t

z

t

z

t

z

z

z

t

t

z

z

z

z

z

z

z























Proof: We consider

. =

) )( )(

( _{1 2 3 1 2 3}

3 2 1 z t z z t y z t x z t z t z t z z z t           

then we have

).

)(

(

)

)(

(

)

)(

(

=

_{2 3 3 1 1 2}

3 2

1

z

z

x

t

z

t

z

y

t

z

t

z

z

t

z

t

z

z

t























where

t

=

z

₁

,

z

₂

,

z

₃ we get

                     ) )( ( = ) )( ( = ) )( ( = 2 3 1 3 2 1 1 2 3 2 1 3 3 1 2 1 3 2 z z z z z z z z z z z z z y z z z z z z x

and we get the result.

Example 9. Given four different complex numbers

4 3 2 1

,

z

,

z

,

z

z

and three complex numbers u,v,w

then we have =

) )( )( )( ( ) )( )( ( 4 3 2

1 x z x z x z

. ) )( )( )( ( ) )( )( ( ) )( )( )( ( ) )( )( ( 4 3 4 2 4 1 4 4 4 4 3 4 3 2 3 1 3 3 3 3 z x z z z z z z w z v z u z z x z z z z z z w z v z u z                

Proof: We consider

. = ) )( )( )( ( ) )( )( ( 4 4 3 3 2 2 1 1 4 3 2

1 x z

x z x x z x x z x x z x z x z x z x w x v x u x              

Then we have

)

)(

)(

(

)

)(

)(

(

)

)(

)(

(

=

)

)(

)(

(

x



u

x



v

x



w

x

₁

x



z

₂

x



z

₃

x



z

₄



x

₂

x



z

₃

x



z

₄

x



z

₁



x

₃

x



z

₄

x



z

₁

x



z

₂

).

)(

)(

(

_{1 2 3}

4

x

z

x

z

x

z

x









Where

4 3 2 1

,

,

,

=

z

z

z

z

x

have

                                   ) )( )( ( ) )( )( ( = ) )( )( ( ) )( )( ( = ) )( )( ( ) )( )( ( = ) )( )( ( ) )( )( ( = 3 4 2 4 1 4 4 4 4 4 2 3 1 3 4 3 3 3 3 3 1 2 4 2 3 2 2 2 2 2 4 1 3 1 2 1 1 1 1 1 z z z z z z w z v z u z x z z z z z z w z v z u z x z z z z z z w z v z u z x z z z z z z w z v z u z x

and we get the result.

Example 10. Given three different complex numbers

z

₁

,

z

₂

,

z

₃ we have

1. = ) )( ( ) )( ( ) )( ( ) )( ( ) )( ( ) )( ( 2 3 1 3 2 1 1 3 1 3 3 2 1 3 3 2 3 2 2 1 3 2 2 1                z z z z z z z z z z z z z z z z z z z z z z z z

Proof: Set = , 2 1 2 1 z z z z x   . = , = 1 3 1 3 3 2 3 2 z z z z z z z z z y    

Then we have

1).

1)(

1)(

(

=

1)

1)(

1)(

(

x



y



z



x



y



z



Therefore we have

xy



yz



zx

=



1

and we get the result.

Example 11. Given three different complex numbers

z

₁

,

z

₂

,

z

₃ we have

)

)(

)(

(

)

)(

)(

(

_{2 3 2 1 2}

2 2 1 3 1 2 1 2 1

z

z

z

z

z

z

z

z

z

z

z

z

z

z















.

)

)(

)(

(

=

)

)(

)(

(

_{1 2 3}

2 3 2 3 1 3 2 3

z

z

z

z

z

z

z

z

z

z

z

z

z

z















Proof: Set

.

=

)

)(

)(

(

₃ 3 2 2 1 1 3 2 1 2

z

z

x

z

z

x

z

z

x

z

z

z

z

z

z

z

















Then we have

).

)(

(

)

)(

(

)

)(

(

=

_{1 2 3 2 3 1 3 1 2}

2

z

z

z

z

x

z

z

z

z

x

z

z

z

z

x

z

















Where

z

=

z

₁

,

z

₂

,

z

₃

we get































)

)(

(

=

)

)(

(

=

)

)(

(

=

2 3 1 3 2 3 3 1 2 3 2 2 2 2 3 1 2 1 2 1 1

z

z

z

z

z

x

z

z

z

z

z

x

z

z

z

z

z

x

and we deduce the result.

Example 12. Given

n



1

complex numbers

z

z

z

z

₁

,

₂

,



,

_n

,

and

z



z

_k

,

k

=

1,2,



,

n

,

we have

. ) )( ( ) )( ( ) )( ( = ) )( ( ₁ 1 3 2 3 2 2 1 2 1 1 1 n n n n n n z z z z z z z z z z z z z z z z z z z z z z z z                  

Proof: From

h k h k h k z z z z z z z z z z     

 1 1

= ) )(

( we

deduce the result.

Example 13. Assume the different complex numbers

z

₁

,

z

₂

,



,

z

_s and z_ia_j 0 where

s

i

=

1,



,

and

j

=

1,2,



,

n

.

Given

x

₁

,

x

₂

,



,

x

_n by

n n n s a z x a z x a z x a z a z a z z z z z z z                2 2 1 1 2 1 2 1 = ) ( ) )( ( ) ( ) )( (

and deduce

. | | | | | | | | | || | | | | | 1 1 1 = 1 1 = 1 = 1 = n s k k i n k i i k k i i k s i n

k z a z a

z z z z a z a a a a z a        









  _   Proof: Assume

.

=

)

(

)

(

)

(

)

(

2 2 1 1 1 1 n n n s

a

z

x

a

z

x

a

z

x

a

z

a

z

z

z

z

z



























then we have

)

(

)

)(

(

)

(

)

)(

(

_{2 3 2 1 3}

1

z

a

z

a

z

a

n

x

z

a

z

a

z

a

n

x



















)

(

)

)(

(

)

(

)

)(

(

_{1 2 4 1 2}

3

z

a

z

a

z

a

n

x

z

a

z

a

z

a

n

x





















)

(

)

)(

(

...





₁



₂



_1



x

_n

z

a

z

a



z

a

_n

).

(

)

)(

(







































































































  

 

n

i n n

i

i n s

i n s

n

i n

i i i

i s

i s

i n

i

i s

i s

i n

i

i s

i s

a

x

a

a

z

a

x

a

x

a

a

a

a

z

a

x

a

x

a

a

a

a

z

a

x

a

x

a

a

z

a

x

=

where

)

(

)

(

1)

(

=

...

=

where

)

(

)

(

)

(

1)

(

=

=

where

)

(

)

(

)

(

1)

(

=

=

where

)

(

)

(

1)

(

=

1

1 =

1 = 1

3

3 4

= 3 2

1 =

3 1 = 2

3

2

2 3

= 1 2

2 1 = 1

2

1

1 2

= 1 1 = 1

and

. ) )( (

) (

) ( 1)

( =

) ( ) (

) ( ) (

1 = 1

1 =

1 = 1

1 = 1

1

k k i n

k i i k k

i

i k s

i k s n

k n s

a z a a a a

z a

a z a z

z z z z

  

 

 

 









 

 

 

Therefore, we deduce

. | | | |

| | | |

| || |

| |

| |

1 1

1 = 1

1 =

1 =

1

= n

s

k k i n

k i i k k

i

i k s

i n

k z a z a

z z z z

a z a a a

a

z a

 

 

  











 

 



3. CONCLUSION

Many teachers believe that they in Vietnam have benefited from innovative general education program after 2015. Teaching method had important innovation in both theory and practical. it is important to teach pupils self-learning and discovering knowledge. teachers help pupils to applying knowledge flexibly and studying more effectively. I and my partner ([HT15, Tin15]) state that, It has been my experience that competency in mathematics, both in complex numbers manipulations and in understanding its conceptual foundations, Enhances a teacher’s ability to teach mathematical in high schools in Vietnam.

REFERENCES

[HT15] T. T. Hai, T. T. Tinh - Forming and strengthening teaching competency in Mathematics Pedagogy students.

Journal of science, Hanoi National of Education, Vol. 60, No. 1, 2015, Vietnam.

[N+15] D. V. Nhi, V. D. Chin, D. N. Dung, P. M. Phuong, T. T. Tinh, N. A. Tuan -

Elementary geometry. Information and Communications Publishing House, 2015, Vietnam.