de Moivre's theorem

**liedora** · July 16th 2011, 08:31 PM

Hi, I'm currently trying to deduce de Moivre's formula,

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

This is what i have so far...

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

Then we have,

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

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.

**chisigma** · July 16th 2011, 09:14 PM

The Euler's identity...

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

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

Kind regards

$\chi$ $\sigma$

**Prove It** · July 17th 2011, 03:31 AM

Originally Posted by liedora

Hi, I'm currently trying to deduce de Moivre's formula,

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

This is what i have so far...

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

Then we have,

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

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

Base step: $\displaystyle n = 2$ .

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

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

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

Q.E.D.

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