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

Hope this helps,
Adam

By Eva on July 31, 1998 :

Dear David,

De Moivre's theorem says that

(cosx + i sinx)ⁿ = cosn x + i sinn x.

This is true for all integers n, and all complex numbers x.

The problem when n is not an integer is that cosn x is single-valued, whereas (cosx + i sinx)ⁿ is many-valued.

Example: x = p/3, n = 1/2. In this case

cosn x + i sinn x = cos(p/6) + i sin(p/6) = Ö3/2 + i/2

whereas (cosx + i sinx)^1/2 = [cos(p/3) + i sin(p/3)]^1/2 = [1/2 + iÖ3/2]^1/2 .

Now [ą(Ö3/2 + i/2)]² = (Ö3/2 + i/2)² = (1/4)(Ö3 + i)² = 1/2 + iÖ3/2.

So we see that

(cosx + i sinx)^1/2 = ą[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),

(ą[cos(x/2) + i sin(x/2)])² = (cosx/2 + i sinx/2)² = cos² (x/2) - sin² (x/2) + 2i sin(x/2) . cos(x/2) = cosx + i sinx.

Thus in general (cosx + i sinx)^1/2 = ą[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 (cosx + i sinx)ⁿ is many-valued (it has infinitely many values if n is irrational), and one of these values is cosn x + i sinn x. This is all that can be said.