De Moivre’s Theorem

 

 

Motivation:

                    Consider the following complex plane shown below for all complex numbers.

 

                    The coordinate for each complex number is composed of it real part and imaginary part.

                           

 

                    If  we draw a circle on this complex plane,  all the points in this unit circle can be represented as the form

               

                                         cosA + i sinA

 

                How about    cos 2A + i sin 2A ? 

                It is also on this unit circle.

               

                But what is the relationship between  (cos 2A + i sin 2A) and (cos A + i sin A) ?

 

Theorem ( De Moivre’s theorem )

                                ( cosA + i sinA )ⁿ = cos(nA) + i sin(nA) ,   n = 1, 2, 3, 4, …

 

              Proof:

 

                                We use mathematical induction approach to prove this theorem.

 

1.      when n=1,  the equality holds trivially:

            LHS= cosA + i sinA  =  RHS

 

2.      Assume  the equality holds when n=k

          (cosA + i sinA)^k = cos (kA) + i sin(kA)

 

3.      Consider the case for n=k+1

                                       LHS = ( cosA + i sinA )^k+1

                                               = ( cosA + i sinA )^k ( cosA + i sinA)

                                               = ( cos (kA) + i sin (kA) ) ( cosA + i sinA )

                                               =  (cos (kA) cosA – sin (kA) sinA)

                                                        + i( sin (kA) cosA + cos (kA) sin A)

                                               =  cos( kA + A ) + i sin( kA + A)

                                               =  cos ( (k+1)A) + i sin ((k+1)A)

                                               = RHS

 

                          The equality holds for n=k+1 under the assumption that it holds for n=k.

 

                            From 1, 2, and 3,  we know that  the equality holds for n=1, 2, 3, 4, 5, …. by induction.

 

 

Application:     Find all the roots of the equation  

                                                               xⁿ + 1 = 0 .

 

                 Sol:

                                          cos  = -1,  sin  = 0.

 

                            Furthermore, 

                                    cos ( 2k  +    ) = -1,  sin ( 2k  +   ) = 0,  k=0, 1, 2, 3, 4, …

                                  

                              Consider     p_k = cos  + i  sin   ,  k = 0, 1, 2, 3, 4, …

 

                                   Then  p_k  satisfies the equation . 

                                   That means p_k is a root of the equation for each k.

 

                            Hence, it is only necessary to find n distinct roots from this sequence p_k .

                                               p_k, k=0, 1, 2, …, (n-1)  are distinct;

 

                               those are all the roots of the equation.