Welcome to NRICH.

 
De Moivre's Theorem: does it apply only to rationals?
By DHChandler (t448) on July 22, 1998:

I understand that DeMoivre's theorem only applies to rational numbers.

My students are not convinced by this and wonder why it cannot apply to any real number ?

By Adam Wood (ajpw2) on July 26, 1998:

Your students are right -- in fact it applies to any complex number, which include the reals.

However I believe de Moivre only proved it for rationals, and many (school) text books only seem to go that far.

The result for complex numbers follows from Euler's Formula:
eix = cos(x) + i sin(x)

Hope this helps,
Adam

By Eva on July 31, 1998:

Dear David,
De Moivre's theorem says that
(cos x + i sin x)n = cos nx + i sin nx.
This is true for all integers n, and all complex numbers x.
The problem when n is not an integer is that cos nx is single-valued, whereas (cos x + i sin x)n is many-valued.

Example: x = p/3, n = 1/2. In this case
cos nx + i sin nx = cos (p/6) + i sin (p/6) = sqr(3)/2 + i/2
whereas (cos x + i sin x)1/2 = [cos(p/3) + i sin (p/3)]½ = [1/2 + isqr(3)/2]½.
Now [±(sqr(3)/2 + i/2)]2 = (sqr(3)/2 + i/2)2 = (1/4)(sqr(3) + i)2 = 1/2 + isqr(3)/2.
So we see that
(cos x + i sin x)1/2 = ±[cos (x/2) + i sin (x/2)].

We have seen this in the case x = p/3, but it is true in general. Indeed, for any x (even complex x),
(± [cos (x/2) + i sin (x/2)])2 = (cos x/2 + i sin x/2)2
= cos2 (x/2) - sin2 (x/2) + 2i sin (x/2) . cos (x/2)
= cos x + i sin x.

Thus in general (cos x + i sin x)1/2 = ±[cos (x/2) + i sin (x/2)].

A similar statement holds for all complex n, and all complex x: if n is not an integer, then (cos x + i sin x)n is many-valued (it has infinitely many values if n is irrational), and one of these values is cos nx + i sin nx. This is all that can be said.