Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(lu) #1

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.

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