Mathematical Methods for Physics and Engineering : A Comprehensive Guide

3.4 DE MOIVRE’S THEOREM

Find an expression forcos^3 θin terms ofcos 3θandcosθ.

Using (3.32),

cos^3 θ=

1

23

(

z+

1

z

) 3

=

1

8

(

z^3 +3z+

3

z

+

1

z^3

)

=

1

8

(

z^3 +

1

z^3

)

+

3

8

(

z+

1

z

)

.

Now using (3.30) and (3.32), we find

cos^3 θ=^14 cos 3θ+^34 cosθ.

This result happens to be a simple rearrangement of (3.29), but cases involving

larger values ofnare better handled using this direct method than by rearranging

polynomial expansions of multiple-angle functions.

3.4.2 Finding thenth roots of unity

The equationz^2 = 1 has the familiar solutionsz=±1. However, now that

we have introduced the concept of complex numbers we can solve the general

equationzn= 1. Recalling the fundamental theorem of algebra, we know that

the equation hasnsolutions. In order to proceed we rewrite the equation as

zn=e^2 ikπ,

wherekis any integer. Now taking thenth root of each side of the equation we

find

z=e^2 ikπ/n.

Hence, the solutions ofzn= 1 are

z 1 , 2 ,...,n=1,e^2 iπ/n, ..., e^2 i(n−1)π/n,

corresponding to the values 0, 1 , 2 ,...,n−1fork. Larger integer values ofkdo

not give new solutions, since the roots already listed are simply cyclically repeated

fork=n, n+1,n+2, etc.

Find the solutions to the equationz^3 =1.

By applying the above method we find

z=e^2 ikπ/^3.

Hence the three solutions arez 1 =e^0 i=1,z 2 =e^2 iπ/^3 ,z 3 =e^4 iπ/^3. We note that, as expected,
the next solution, for whichk=3,givesz 4 =e^6 iπ/^3 =1=z 1 , so that there are only three
separate solutions.