Abraham de Moivre and formula

De Moivre’s formula states that (cos x + isin x)^n = cos nx + isin nx.

Where i is the imaginary number √-1.

Now, for proof and simplicity’s sake, let’s change cosx + isin x to cis x.

This is a proof by induction, as stated by Wikipedia (I use a lot of Wikipedia these days.)

Let the statement (cos x + isin x)^n = cos nx + isin nx be S(n) for n an integer.

For n>0, it is clear that S(1) is true. Now, assume that S(k) is tru for a natural k, or (cis x)^k = cis kx

Consider S(k + 1):

(cis x)^(k + 1) = (cis x)^k * cis x = (cis kx)(cis x)† = (cos kx)(cos x) – (sin kx)(sin x) + i{(cos kx)(sin x) + (sin kx)(cos x)} = cis {(k + 1)x}††

† by the induction hypothesis

†† by the trigonometric identities

Because we have proven that (cis x)^(k + 1) = cis {(k+1)x), we can deduce that S(n) implies S(k + 1).

For S(0), it is clearly true since cos 0 + isin 0 = 1 and any number to the power of 0 (except zero itself) is 1.

For the negatives, let the exponent be -n for a natural n.

(cis x)^-n = (cis x)^n^-1 = (cis nx)^-1 = cos -nx + isin -nx

I don’t know what this is about, but the bold equation is a result of the identity

{\displaystyle z^{-1}={\frac {\bar {z}}{|z|^{2}}},}

where z = cis nx.

Therefore, S(n) holds for any integer n.

QED and for a better understanding, go check out link.

2 thoughts on “Abraham de Moivre and formula

  1. I started this in the morning yesterday, but I didn’t get to finish this, so that’s why I couldn’t post yesterday. I’m trying to post as much as I can, but not too much, so maybe daily, or at the slowest, every two days. I also plan on deleting some parts of the navigation bar. All for today, I guess.


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