de Moivre's theorem

• Jul 16th 2011, 08:31 PM
liedora
de Moivre's theorem
Hi, I'm currently trying to deduce de Moivre's formula,

$\displaystyle (\cos{\theta} + i\sin{\theta})^n = \cos{n\theta} + i\sin{n\theta}$.

This is what i have so far...

If $\displaystyle z = \cos{\theta} + i\sin{\theta}$, $\displaystyle \arg{z} = \theta$ and $\displaystyle \arg{z^n} = n\theta$

Then we have,

$\displaystyle z^n = r^n(\cos{\theta}+i\sin{\theta})^n\\= (\cos{\theta}+i\sin{\theta})^n\\=\cos(\arg(z^n))+i \sin(arg(z^n))\\=\cos{n\theta} + i\sin{n\theta}$

But I can't seem to figure out how they got from the second to the third step, It would be greatly appreciated if someone could please explain this.
• Jul 16th 2011, 09:14 PM
chisigma
Re: de Moivre's theorem
The Euler's identity...

$\displaystyle e^{i \theta} = \cos \theta + i\ \sin \theta$ (1)

... combined with elementary properties of exponential conducts immediately to the result...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jul 17th 2011, 03:31 AM
Prove It
Re: de Moivre's theorem
Quote:

Originally Posted by liedora
Hi, I'm currently trying to deduce de Moivre's formula,

$\displaystyle (\cos{\theta} + i\sin{\theta})^n = \cos{n\theta} + i\sin{n\theta}$.

This is what i have so far...

If $\displaystyle z = \cos{\theta} + i\sin{\theta}$, $\displaystyle \arg{z} = \theta$ and $\displaystyle \arg{z^n} = n\theta$

Then we have,

$\displaystyle z^n = r^n(\cos{\theta}+i\sin{\theta})^n\\= (\cos{\theta}+i\sin{\theta})^n\\=\cos(\arg(z^n))+i \sin(arg(z^n))\\=\cos{n\theta} + i\sin{n\theta}$

But I can't seem to figure out how they got from the second to the third step, It would be greatly appreciated if someone could please explain this.

You can use induction...

You wish to show $\displaystyle \displaystyle (\cos{\theta} + i\sin{\theta})^n = \cos{n\theta} + i\sin{n\theta}$.

Base step: $\displaystyle \displaystyle n = 2$.

\displaystyle \displaystyle \begin{align*} (\cos{\theta} + i\sin{\theta})^2 &= \cos^2{\theta} + 2i\sin{\theta}\cos{\theta} + i^2\sin^2{\theta} \\ &= \cos^2{\theta} - \sin^2{\theta} + 2i\sin{\theta}\cos{\theta} \\ &= \cos{2\theta} + i\sin{2\theta} \end{align*}

Inductive step: Assume the statement is true for $\displaystyle \displaystyle n = k$, in other words, assume $\displaystyle \displaystyle (\cos{\theta} + i\sin{\theta})^k = \cos{k\theta} + i\sin{k\theta}$, then we need to show the statement is true for $\displaystyle \displaystyle n = k + 1$, in other words, that $\displaystyle \displaystyle (\cos{\theta} + i\sin{\theta})^{k+1} = \cos{(k + 1)\theta} + i\sin{(k + 1)\theta}$.

\displaystyle \displaystyle \begin{align*}(\cos{\theta} + i\sin{\theta})^{k + 1} &= (\cos{\theta} + i\sin{\theta})^k(\cos{\theta} + i\sin{\theta}) \\ &= (\cos{k\theta} + i\sin{k\theta})(\cos{\theta} + i\sin{\theta}) \\ &= \cos{k\theta}\cos{\theta} + i\cos{k\theta}\sin{\theta} + i\cos{\theta}\sin{k\theta} + i^2\sin{k\theta}\sin{\theta} \\ &= \cos{k\theta}\cos{\theta} - \sin{k\theta}\sin{\theta} + i(\cos{k\theta}\sin{\theta} + \cos{\theta}\sin{k\theta}) \\ &= \cos{(k\theta + \theta)} + i\sin{(k\theta + \theta)}\\ &= \cos{(k + 1)\theta} + i\sin{(k + 1)\theta} \end{align*}

Q.E.D.