Math, asked by skkamrulislam364, 4 months ago

state demoivers theorem and apply it to solve x^6-3x^3+3=0​

Answers

Answered by aaditamodas
0

Answer:

If

z

3

1

=

0

, then we are looking for the cubic roots of unity, i.e. the numbers such that

z

3

=

1

.

If you're using complex numbers, then every polynomial equation of degree

k

yields exactly

k

solution. So, we're expecting to find three cubic roots.

De Moivre's theorem uses the fact that we can write any complex number as

ρ

e

i

θ

=

ρ

(

cos

(

θ

)

+

i

sin

(

θ

)

)

, and it states that, if

z

=

ρ

(

cos

(

θ

)

+

i

sin

(

θ

)

)

, then

z

n

=

ρ

n

(

cos

(

n

θ

)

+

i

sin

(

n

θ

)

)

If you look at

1

as a complex number, then you have

ρ

=

1

, and

θ

=

2

π

. We are thus looking for three numbers such that

ρ

3

=

1

, and

3

θ

=

2

π

.

Since

ρ

is a real number, the only solution to

ρ

3

=

1

is

ρ

=

1

. On the other hand, using the periodicity of the angles, we have that the three solutions for

θ

are

θ

1

,

2

,

3

=

2

k

π

3

, for

k

=

0

,

1

,

2

.

This means that the three solutions are:

ρ

=

1

,

θ

=

0

, which is the real number

1

.

ρ

=

1

,

θ

=

2

π

3

, which is the complex number

1

2

+

3

2

i

ρ

=

1

,

θ

=

4

π

3

, which is the complex number

1

2

3

2

i

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