## Thursday, December 27, 2007

### Euler's Proof of De Moivre Formula

De Moivre's Formula is easily proven using induction. The following proof was first given by Leonhard Euler in 1748. Abraham De Moivre never stated his formula exactly in this form.

Lemma 1: (cos α + isin α)(cos β + isin β) = cos(α + β) + isin(α + β)

Proof:

(1) cos(α + β) = cos(α)cos(β) - sin(α)sin(β) [See Theorem 2, here]

(2) sin(α + β) = cos(α)sin(β) + cos(β)sin(α) [See Theorem 1, here]

(3) isin(α + β) = cos(α)isin(β) + cos(β)isin(α)

(4) Since isin(α)*isin(β) = -sin(α)sin(β), we have:

cos(α + β) + isin(α + β) =

= cos(α)cos(β) - sin(α)sin(β) + cos(α)isin(β) + cos(β)isin(α) =

= cos(α)[cos(β) + isin(β)] + isin(α)[cos(β) + isin(β)] =

= [cos(α) + isin(α)][cos(β) + isin(β)]

QED

Theorem 1: De Moivre's Formula

(cos α + i sin α)n = cos(nα) + isin(nα) for all α ∈ R.

Proof:

(1) For n=1:

(cos α + i sin α)1 = cos(1*α) + isin(1*α)

(2) Assume that the premise is true for all values up to n such that:

(cos α + i sin α)n = cos(nα) + isin(n&alpha) for all α ∈ R.

(3) (cos α + isin α)n+1 =
(cos α + isin α)n(cos α + isin α)

(4) Using the assumption in step #2, we have:

(cos α + isin α)n(cos α + isin α) =

[cos(n α) + isin(n α)](cos α + isinα)

(5) Using Lemma 1 above, we have:

[cos(n α) + isin(n α)](cos α + isinα) =

cos(n α + α) + isin(nα + α) =

= cos ([n+1]α) + isin([n+1]α)

QED

References

#### 1 comment:

Amateur said...

I need some help.
I have to find examples of different problems easily solved using De Moivre Formula or Roots of Unity (geometry, trig., De Moivre Formula together with Binomial Formula???).
Any piece of advice would be usefull.
As soon as possible. Literally.