A First Course in Linear Algebra, 2017a

334 Complex Numbers
This requires r k = |z| and so r = |z| 1/k . Also, both cos(kα) =cosθ and sin(kα)=sinθ. This can only
happen if
kα = θ + 2lπ
for l an integer. Thus
α = θ + 2lπ , l = 0,1,2,···,k − 1
k
and so the k th roots of z are of the form
( ) ( ))
|z|
(cos
1/k θ + 2lπ θ + 2lπ
+ isin
, l = 0,1,2,···,k − 1
k
k
Since the cosine and sine are periodic of period 2π, there are exactly k distinct numbers which result from
this formula.
♠
The procedure for finding the kk th roots of z ∈ C is as follows.
Procedure 6.17: Finding Roots of a Complex Number
Let w be a complex number. We wish to find the n th roots of w, thatisallz such that z n = w.
There are n distinct n th roots and they can be found as follows:.
1. Express both z and w in polar form z = re iθ ,w = se iφ .Thenz n = w becomes:
We need to solve for r and θ.
2. Solve the following two equations:
(re iθ ) n = r n e inθ = se iφ
r n = s
e inθ = e iφ (6.1)
3. The solutions to r n = s are given by r = n√ s.
4. The solutions to e inθ = e iφ are given by:
nθ = φ + 2πl, for l = 0,1,2,···,n − 1
or
θ = φ n + 2 πl, for l = 0,1,2,···,n − 1
n
5. Using the solutions r,θ to the equations given in (6.1) construct the n th roots of the form
z = re iθ .
Notice that once the roots are obtained in the final step, they can then be converted to standard form
if necessary. Let’s consider an example of this concept. Note that according to Corollary 6.16, thereare
exactly 3 cube roots of a complex number.