MATH 117
The Roots of Complex
Numbers
Imaginary numbers were discovered while attempting to evaluate the square roots of
negative numbers within the context of attempting to solve the depressed cubic
equation. This discovery led to the initial definition of the imaginary number i = −1 .
Square roots of other negative numbers then could be defined such as −9 = 3 i .
We can now solve equations that previously had no real solutions. For example if
x + 9 = 0, then x 2 = −9 and x = ± −9 = ± 3i . The polynomial x 2 + 9 then factors as
(x + 3 i)(x − 3i) .
2
4
What about the equation x + 9 = 0? How many solutions will there be? In
n
general, what are the solutions to x = a ? This general equation requires that we
compute the n th root of the real number a . Because n could be even or odd and a
could be positive or negative, there seems to be four possible cases. However we need
complex numbers to compute n a only when n is even and a is negative.
n
On the other hand, x = a has more solutions than just x = n a . There are actually
n distinct complex solutions to this equation. How do we find all such solutions? Can
n
we similarly solve x = z , where z is a complex number?
The Polar Form Solution
n
To solve the equation x = z , first recall the forms in which the complex number z may
be written:
z = a + bi = r e i θ = r (cos θ + i sin θ ) ,
where r is a real number with r ≥ 0, and θ is an angle such that 0 ≤ θ < 2π.
It is especially important that we initially write θ as an angle between 0 and 2π. By
adding multiples of 2π, we obtain angles that are co-terminal to θ and therefore have
the same cosine and sine. In other words we obtain the same complex number
z = r e i θ = r e i (θ + 2k π) ,
for all integers k , in particular for k = 0, 1, 2, . . . , n − 1 .
Now for 0 ≤ θ < 2 π and 0 ≤ k ≤ n − 1 ,
0 ≤ θ + 2 k π < 2 π + 2 k π = 2 π (k + 1) ≤ 2 nπ .
θ + 2k π
< 2 π . As k varies, there are n
n
distinct angles described here which are equally spaced around the unit circle between
0 and 2π.
Thus for k = 0, 1, 2, . . . , n − 1 , we have 0 ≤
Because the radius r is a non-negative real number, the value n r is defined. So
consider the n distinct complex numbers
 θ + 2 k π 

n
 θ + 2 kπ
 θ + 2k π   n i 
+
i
sin
zk = n r cos


 = r e

n
n
,
for k = 0, 1, 2, . . . , n − 1 . Then
n

 θ + 2 k π  
i
(zk ) n =  n r e  n   = r e i (θ + 2k π) = z .


n
Thus there are n distinct complex solutions to the equation x = z given by the
values of zk for k = 0, 1, 2, . . . , n − 1 . These solutions will be equally spaced around a
circle of radius n r .
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Example 1. Find all complex solutions to the equation x = 16+ 16 3 i .
Solution. We first write 16+ 16 3 i in polar form. Here r =
162 + (16 3 )2 = 32. Also
tan −1 (b / a) = tan −1 ( 3) = π/3. So θ = π/3 since the point (16, 16 3 ) is in the 1st
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quadrant. So z = 32 ( cos(π / 3) + i sin(π / 3)) . The five solutions to x = z are
  π / 3 + 2k π 
 π / 3 + 2 kπ
zk = 5 32 cos
+ i sin



 
5
5
  π 2 kπ
 π 2k π  
= 2 cos +
+ i sin  +


15
5
15
5 
( (
))
= 2 cos 12 + 72 k  + i sin 12 + 72 k  ,
)
(
for k = 0, 1, 2, 3, 4.
We have written the final form in degrees so as to better visualize the result.
To compute the rectangular form of the solutions, we simply let k vary from 0 to 4
and evaluate. The five solutions are equally spaced around a circle of radius 2.
z0 = 2 cos(12 ) + i sin(12 ) ≈ 1.956295 + 0.4158234 i .
(
)
z1 = 2 cos(84 ) + i sin(84 ) ≈ 0.209 + 1.989 i
(
)
z2 = 2 cos(156 ) + i sin(156 ) ≈ −1.827 + 0.81347 i
(
)
z3 = 2 cos(228 ) + i sin(228 ) ≈ −1.33826 − 1.4863 i
(
)
z4 = 2 cos(300 ) + i sin(300 ) ≈ 1− 1.732 i
(
)
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Example 2. Find all complex solutions to the equation x = –256.
Solution. Because –256 is the point (–256, 0), the polar coordinates are r = 256 and θ =
π. The four solutions to the equation are
  π + 2k π 
 π + 2 k π 
zk = 4 256 cos
+ i sin





4
4
  π kπ
 π kπ
= 4 cos +
+ i sin +


4
2
4
2 
( (
= 4 cos 45 + 90 k  + i sin 45 + 90k 
)
(
)) ,
for k = 0, 1, 2, 3. The rectangular form of the solutions are
z0 = 4 cos(45 ) + i sin(45 ) = 2 2 + 2 2 i
(
)


z1 = 4 (cos(135 ) + i sin(135 ) ) = −2 2 + 2 2 i
z2 = 4 ( cos(225 ) + i sin(225 )) = −2 2 − 2 2 i
z3 = 4 (cos(315 ) + i sin(315 )) = 2 2 − 2 2 i
Notice that the pairs ( z0 , z3 ) and ( z1 , z2 ) are conjugates having the form a ± b i .
When a polynomial equation has real coefficients, then it is always the case that the
complex solutions occur in conjugate pairs. Thus consider the following four cases:
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(i) x = 100
6
(ii) x = –100
7
(iii) x = 100
7
(iv) x = –100
Case (i): There are two real solutions ± 6 100 and four complex solutions occurring in
two complex conjugate pairs.
Case (ii): There are no real solutions, but six complex solutions occurring in three
complex conjugate pairs.
Case (iii): There is one real solution 7 100 and six complex solutions occurring in three
complex conjugate pairs.
Case (iv): There is one real solution – 7 100 and six complex solutions occurring in three
complex conjugate pairs.
These cases can be generalized for all (positive) even and odd exponents.
Exercises
Find all complex solutions to the equations
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(a) x = 15625 i
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(b) x = –2.48832
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(c) x = −648 + 648 3 i