Roots of Complex Numbers
Section 9.3
nth root
• If z and w are complex
numbers and if n ≥ 2 is
an integer, then z is an
nth root of w iff zn = w
Example 1
• Write the 3 cube roots of 1331 in
trigonometric form.
First, z3 = 1331
z = [r, θ]
by De Moivre’s
z3 = [r3, 3θ] = [1331, 0]
So r3 = 1331 so, r = 11
3θ = 0 + 2π
or θ = 0 + 2/3π
Example 1
3θ = 0 + 2π
So, we have
or θ = 0 + 2/3π
[11, 0]  11 (cos (0) + isin (o))  11
[ 11, 2/3 π]  11 (cos (2π/3)+ isin (2π/3))
[11, 4/3 π]  11 (cos (4π/3)+ isin (4π/3)
nth root of complex numbers
theorem
• The n nth roots of [r, θ] are
2
[ r,  k ]
where k = 0, n
1, 2, .., n-1n
n

The n nth roots of r(cos θ + isin θ)

2

2
n
r (cos(  k
)  i sin(  k
))
n
n
n
n
Where k = 0, 1, 2, …, n-1
Example 2
Express the sixth roots of [729, 60˚]
Nth roots of complex numbers theorem
60
360
[ 729 ,  k 
]
6
6
6
[3, 10˚] [3, 70˚]
[3, 250˚]
[3, 130˚] [3, 190˚]
[3, 310˚]
Example 3
Express the square roots of i in polar,
trigonometric, and binomial.
Rectangular: (0,1)
Polar: [1, π/2]
 /2
2
2
k ]
Square roots in polar: [ 1 ,

[1, ]
4
2
5
[1, ]
4
2
Example 3
 
 
[1, ] to trigonometric: 1(cos   i sin )
4
4
4
2
2
to binomial:
i
2
2
5
5

5




[1, ] to trigonometric: 1(cos   i sin )
4
 4 
 4 

to binomial:
2
2

i
2
2
Example 4
There are exactly two real nth roots of a positive
number if n is even and exactly one real nth root if
n is odd. The other nth roots are vertices of a
regular n-gon centered at the origin.
Example: z5 = 51 Regular pentagon
What figure does this make?
How many real and nonreal solutions does this
make?
Real: 1
Nonreal: 4
Homework
Pages 539 – 540
2, 4 – 10, 13