Complex Numbers Worksheet - Appendix F, Cengage Page 9
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APPENDIX F
Complex Numbers
To find a formula for an nth root of a complex number, let u be an nth root of z,
The nth roots of a
STUDY TIP
complex number are useful for solving
where
some polynomial equations. For
u
s cos
i sin
and
z
r cos
i sin
.
instance, you can use DeMoivre’s
Theorem to solve the polynomial
n
By DeMoivre’s Theorem and the fact that
u
z,
you have
equation
n
s
cos n
i sin n
r cos
i sin
.
4
x
16
0
n
Taking the absolute value of each side of this equation, it follows that
s
r.
by writing
16
as 16 cos
i sin
.
Substituting back into the previous equation and dividing by r, you get
cos n
i sin n
cos
i sin .
So, it follows that
cos n
cos
and
sin n
sin .
Because both sine and cosine have a period of
2 ,
these last two equations have
solutions if and only if the angles differ by a multiple of
2 .
Consequently, there must
exist an integer k such that
n
2 k
2 k
.
n
By substituting this value for
into the polar form of u, you get the following result.
THEOREM F.2
nth Roots of a Complex Number
For a positive integer n, the complex number
z
r cos
i sin
has exactly n
distinct nth roots given by
2 k
2 k
n
r cos
i sin
n
n
Imaginary
axis
where k
0, 1, 2, . . . , n
1.
2 π
When k exceeds
n
1,
the roots begin to repeat. For instance, if
k
n,
the angle
n
n
r
2 π
2 n
n
Real
2
axis
n
n
is coterminal with
n,
which is also obtained when
k
0.
This formula for the nth roots of a complex number z has a nice geometric
interpretation, as shown in Figure F.4. Note that because the nth roots of z all have
n
n
the same magnitude
r,
they all lie on a circle of radius
r
with center at the origin.
Furthermore, because successive nth roots have arguments that differ by
2
n,
the n
Figure F.4
roots are equally spaced along the circle.
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