Understanding Engineering Mathematics

Problem 12.13

Evaluate

(

2 kp

3

)

fork=0, 1, 2, 3.

Substituting forkwe find:

k= 0 ,^

(

2 kπ

3

)

=^ ( 0 )= 1

k= 1 ,^

(

2 π

3

)

=cos

(

2 π

3

)

+jsin

(

2 π

3

)

=−

1

2

+j

√

3

2

k= 2 ,^

(

4 π

3

)

=cos

(

4 π

3

)

−jsin

(

4 π

3

)

=−

1

2

−j

√

3

2

k= 3 ,^

(

6 π

3

)

=^ ( 2 π)= 1

We s e e t h a tk=3 brings us back to where we started. You can soon convince yourself
that higher values ofksimply reproduce the three roots in the order ofk= 0 , 1 ,2.
So, we obtain all the different roots in polar form by first generalising the angleθto
θ+ 2 kπ, taking the root and then lettingktake the appropriate number of values in turn,
usually commencing withk=0. Problem 12.13 should now help you to appreciate the
general situation, which we now summarise.
We can obtain allqqth roots of cosθ+jsinθby writing it in its most general form:

cosθ+jsinθ≡cos(θ+ 2 kπ)+jsin(θ+ 2 kπ)

k= 0 ,± 1 ,± 2 ,...

Then
[cos(θ+ 2 kπ)+jsin(θ+ 2 kπ)]^1 /q

=cos

(

θ+ 2 kπ

q

)

+jsin

(

θ+ 2 kπ

q

)

k= 0 , 1 , 2 ,...,(q− 1 )

and by lettingktake anyqconsecutive values, for example,k= 0 , 1 , 2 ,...(q− 1 ),we
obtain all of theqqth roots of cosθ+jsinθ.
Theqqth roots of a complex numberz=x+jycan be obtained by converting to
polar form and then using De Moivres theorem:

z^1 /q≡[r(cosθ+jsinθ)]^1 /q

=r^1 /q

[

cos

(

θ+ 2 kπ

q

)

+jsin

(

θ+ 2 kπ

q

)]

k= 0 , 1 , 2 ,...,(q− 1 )

In geometrical terms these roots must all lie at equally spaced intervals of

2 π

q

around the

circle radiusr^1 /q, centre the origin in the complex plane.