De Moivre's formula

Question

I'm starting to study complex numbers, obviously we've work with De Moivre's formula. I was courious about the origin of it and i look for the original paper, I found it in the Philosophicis Transactionibus Num. 309, "De sectione Anguli", but only in latin, so some words are difficult to understand, however, on the math part I don't see where's the formula.

Answer 1

I always found De Moivre's theorem quite intuitive. I'll try and explain why (I know this isn't what you asked for but it might give you an idea why such a formula was thought to exist).

Think about multiplying two complex numbers $z_1, z_2$. You can do it algebraically in the $a+ib$ form and get a formula or you can do it with the modulus-argument form $r(\cos\theta + i \sin\theta)$.

If you do it in the modulus argument form you find, after a bit of trig manipulation, that the effect of this multiplication is to add the arguments and to multiply the moduli, i.e. if the arguments are $\theta_1, \theta_2$ and the moduli are $r_1, r_2$ then the product is $r_1r_2(\cos(\theta_1+\theta_2) + i \sin(\theta_1+\theta_2))$. Geometrically this is doing a rotation of one by the other and a stretch/squash outwards/inwards.

Well now think about what happens if you square a complex number $z$. In this case $\theta_1=\theta_2=\theta$ and $r_1=r_2=r$ so when squaring you are going to get a complex number with double the argument and square of the original modulus, i.e. $z^2 = z.z = r^2(\cos(2\theta)+i \sin(2\theta))$.

What about if you cube? You can see that you will get $z^3=z.z^2 = r^3(\cos(3\theta)+i \sin(3\theta))$ and intuitively you can see what is happening each time. Inductively you then find that $z^n = r^n(\cos(n\theta)+i \sin(n\theta))$ which you notice is De Moivre's theorem.

Answer 2

DeMoivre only suggested the formula in $1722$. It was Euler who proved it in $1749$. Later DeMoivre's formula was discovered not only for complex numbers, but also for quaternions.