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