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Proving De Moivre's theorm

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Heinsbergrelatz

Posted: Sat Mar 06, 2010 1:27 am Post subject: Proving De Moivre's theorm

Forum Masters Degree

Joined: 01 Aug 2009
Posts: 598
Location: Singapore

can anyone help me prove this?; thanks

$(|z|cis\theta)^{2}=|z|^{n}cisn\theta$

thankyou in advance Very Happy

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$G_\mu_\nu + \Lambda g_\mu_\nu=\frac{8\pi G}{c^{4}} T_\mu_\nu$

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Heinsbergrelatz

Posted: Sat Mar 06, 2010 1:32 am Post subject:

Forum Masters Degree

Joined: 01 Aug 2009
Posts: 598
Location: Singapore

oo wait does this look right? i think i have one way of proving, but im not sure if its correct.

$z=|z|cis\theta$
$z^{2}=|z|cis\theta \times |z|cis\theta$
$=|z|^{2}cis(\theta+\theta)$
$=|z|^{2}cis2\theta$

if its wrong, please correct me
_________________
------------------

$G_\mu_\nu + \Lambda g_\mu_\nu=\frac{8\pi G}{c^{4}} T_\mu_\nu$

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DrRocket

Posted: Sat Mar 06, 2010 6:45 am Post subject: Re: Proving De Moivre's theorm

Forum Radioactive Isotope

Joined: 12 Aug 2008
Posts: 3114

Heinsbergrelatz wrote:

can anyone help me prove this?; thanks

$(|z|cis\theta)^{2}=|z|^{n}cisn\theta$

thankyou in advance Very Happy

$cis(\theta)= e^{i \theta}$ This is called "Euler's formula and is one of the more important relations in all of mathematics. It follows from the definition of the exponential function as a power series.

So for any complex number $z. z=|z|e^{i \theta}$ for some $\theta$ commonly called the argument of

.

From this it follows that $z^n = |z|^n (e^{i\theta})^n = |z|^n e^{in \theta}$

or $(|z|cis\theta)^{n}=|z|^{n}cisn\theta$

You can also prove this by induction on n using trigonometric identies, which is how it is often done in elementary algebra texts.

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