Proving De Moivre's Theorem

Date: 12/03/97 at 08:42:19
From: Chris Pountney
Subject: De Moivre's theorem

Dear Dr. Math,

Please can you show me how to prove De Moivres theorem :

     (cos(x)+isin(x))^n = cos(nx) + isin(nx)

Thank you, 
Chris

Date: 12/08/97 at 10:34:38
From: Doctor Bruce
Subject: Re: De Moivres theorem

Hello Chris,

I'm assuming that your 'n' is meant to represent an integer.  
DeMoivre's Theorem is true even if  n  is a complex number (has a real 
part and possibly an imaginary part), but when  n  is an integer we 
can prove the formula easily by using some basic trigonometry.

We will prove DeMoivre's Theorem for any positive integer  n  by 
induction.

First, the theorem is clearly true for n = 1, since then both sides 
of the equation are the same quantity. So now we will assume that the
theorem has been shown true for n = k. We want to show that it is 
true for  n = k + 1.

We calculate:

  [cos(x) + i sin(x)]^(k+1) = 
  [cos(x) + i sin(x)]^k * [cos(x) + i sin(x)]  (split up factors)

        =  [cos(kx) + i sin(kx)] * [cos(x) + i sin(x)]
           (induction hypothesis)

        =  cos(kx)cos(x) - sin(kx)sin(x) + i[sin(kx)cos(x) + 
           cos(kx)sin(x)] (multiply out)

        =  cos(kx + x) + i sin(kx + x) (use trig formulas)

        =  cos((k+1)x) + i sin((k+1)x).
           (add the angles)

This proves the inductive step, so DeMoivre's Theorem has been shown 
to be true for all positive integers  n.

To show the theorem is also true for negative integers, we just 
compute (still assuming that  n  itself is positive)

     [cos(x) + i sin(x)]^(-n)  =  [{cos(x) + i sin(x)}^(-1)]^n

                               =  [cos(x) - i sin(x)]^n

                               =  [cos(-x) + i sin(-x)]^n

                               =  cos(-nx) + i sin(-nx).

So DeMoivre's Theorem is true for all integers, positive and negative.

I don't know if you know this already, but both the sine and cosine 
functions can be expressed in terms of the exponential function.  
Namely,

     cos(x)  =  [e^(ix) + e^(-ix)]/2
     sin(x)  =  [e^(ix) - e^(-ix)]/2i,

from which we get

     e^(ix)  =  cos(x) + i sin(x).

This is true for any complex number x. This is the usual way we state 
DeMoivre's formula. We see that

     [cos(x) + i sin(x)]^n  =  [e^(ix)]^n    and

     cos(nx) + i sin(nx)    =  e^(inx),

and the two righthand sides above are clearly equal. So this an easier 
way of proving the theorem that you stated, although first you have to 
know about the exponential function.

A fundamental and truly amazing fact that comes out of this is

     e^(i*pi)  =  -1,

which you get by substituting  x  =  pi  in DeMoivre's formula.

-Doctor Bruce,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/