I have a book that has a brief history of the complex numbers and it covers de Moivre's formula:

$(\cos(x) + i\sin(x))^n = \cos(nx) + i\sin(nx)$.

I am very curious as to how this result was originally found, or derived, BEFORE Euler's formula was around. Also, what was the original proof of this?

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    $\begingroup$ Probably by induction. $\endgroup$ – dustin Feb 6 '15 at 1:46
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    $\begingroup$ Induction indeed...and basic trigonometric formulas, of course. $\endgroup$ – Timbuc Feb 6 '15 at 1:52
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    $\begingroup$ @Christopher have answer working on putting it together. $\endgroup$ – dustin Feb 6 '15 at 2:06
  • $\begingroup$ I was puzzled by this result when I saw it because the way I was introduced to multiplication of complex numbers was that one multiplies the absolute values and adds the angles, so this formula seemed like a complete triviality. $\endgroup$ – Michael Hardy Feb 6 '15 at 2:41

Euler was the first to write down de Moirve's formula which I found in The Elementary Mathematical Works of Leonhard Euler starting on page 29. Euler found $$ (\cos(x)\pm i\sin(x))(\cos(y)\pm i\sin(y))(\cos(z)\pm i\sin(z)) = \cos(x \pm y \pm z)\pm i\sin(x\pm y\pm z) $$ from here he deduce what we call de Moirve's theorem $$ (\cos(x)\pm i\sin(x))^n = \cos(nx)\pm i\sin(nx) $$ Then it follows that \begin{align} \cos(nx) &= 1/2[(\cos(x)+ i\sin(x))^n+(\cos(x) - i\sin(x))^n]\tag{1}\\ \sin(nx) &= 1/2[(\cos(x)+ i\sin(x))^n-(\cos(x) - i\sin(x))^n]\tag{2} \end{align} Euler than expanded equations $(1)$ and $(2)$ by the binomial theorem. He then let $x$ be a small angle approximation and $n$ be infinitely large so that $xn$ was finite. Let $xn=v$. Then $\sin(x) = x = v/n$ and we get \begin{align} \cos(v) &= 1 - \frac{v^2}{2!} + \cdots\\ \sin(v) &= v - \frac{v^3}{3!} + \cdots \end{align} Using equations $(1)$ and $(2)$ and letting $j = n$ be the infinitely large number so that $jx=v$ so that $nx = v = v/j$. Now we have $\sin(x) = v/j$ and $\cos(x)=1$. With these substitutions, we get \begin{align} \cos(v) &=\frac{(1+iv/j)^j+(1-iv/j)^j}{2}\\ \sin(v) &=\frac{(1+iv/j)^j-(1-iv/j)^j}{2i} \end{align} In the linked document, the author who compiled the document (Euler) previously has shown that $(1+z/j)^j = e^z$. Letting $z=\pm iv$, we get \begin{align} e^{iv} &= \cos(v) + i\sin(v)\\ e^{-iv} &= \cos(v) - i\sin(v) \end{align}

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    $\begingroup$ Great answer and great link! Thank you so much. $\endgroup$ – Christopher Feb 6 '15 at 2:39
  • $\begingroup$ @Christopher you probably want to go through the link since I left out the some details as well as the binomial expansion of $(1)$ and $(2)$. $\endgroup$ – dustin Feb 6 '15 at 2:39
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    $\begingroup$ @dustin Very nice approach! $\endgroup$ – Math-fun Feb 16 '15 at 20:49
  • $\begingroup$ @Mehdi unfortunately, Euler gets all the credit here. I have no idea why this was named after de Moirve though. It mentions he was working on some equations but I don't see the relation or why he gets the credit. $\endgroup$ – dustin Feb 16 '15 at 20:50
  • $\begingroup$ @dustin well he has been great (and is). By the way I just found an interesting youtube video on Euler: youtube.com/watch?v=h-DV26x6n_Q $\endgroup$ – Math-fun Feb 16 '15 at 20:54

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