but two of them are complex. The question is, how do we find such roots in general?
Since the polar form is so useful in multiplication, and De Moivre’s theorem is useful in
raising to a power, we suspect that we can also use this form in taking roots. We can, but
there is a subtlety that takes some getting used to.
Consider the cube root 11(^3). Written in polar form we have
1
1
(^3) =(cos 0+jsin 0)
1
(^3) =(^ ( 0 ))
1
3
Applying De Moivre’s theorem as it would work for an integer gives
1
1
(^3) =
(
1
3
0
)
=^ ( 0 )= 1
i.e. it only gives us the root 1. But we have missed a trick, namely that because over
cos(θ+ 2 kπ)=cosθand sin(θ+ 2 kπ)=sinθfor any integerk, we can always write
(^) (θ)= (^) (θ+ 2 kπ) k=any integer
as we noted in Section 12.3. That is, we can increaseθby any integer multiple of 2π,and
we won’t change^ (θ). Normally, we wouldn’t need to do this – we simply reproduce the
same values of^ (θ). However, when we take roots, the following happens:
(^ (θ))
1
(^3) =( (^) (θ+ 2 kπ))
1
3
=^
(
θ+ 2 kπ
3
)
when we apply De Moivre’s theorem.Now
(
θ
3
)
=^
(
θ
3
- 2 kπ
3
)
for all integer values ofk
For example, withk=1:
(
θ
3
)
=^
(
θ
3
2 π
3
)
Applying this extension of De Moivre’s theorem to 1=^ ( 0 )gives
(^ ( 0 ))
1
(^3) =[^ ( 0 + 2 kπ)]
1
(^3) k= 0 ,± 1 ± 2 ,...
=^
(
0 + 2 kπ
3
)
=^
(
2 kπ
3
)
k= 0 ,± 1 ,± 2 ,...
Nowwe appear to havetoo manyvalues – one for each of the infinite number of values
ofk. However, you will find that these all reduce to just three different values: