squaring exponentials

**adam_leeds** · March 24th 2011, 12:04 PM

is $(e^(i \theta))(e^(i \theta))$ = $e^(2i \theta))$

**adkinsjr** · March 24th 2011, 12:18 PM

Yes, it is true. You have the two formulae:

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

For $2\theta$

$e^{i(2\theta)}=cos(2\theta)+isin(2\theta)$

You can show that your equation is true as follows:

$(e^{i\theta})(e^{i\theta})$

$=\left[cos(\theta)+isin(\theta)\right]\left[cos(\theta)+isin(\theta)\right]$

$=cos^2(\theta)+2isin(\theta)cos(\theta)+i^2sin^2(\ theta)$

Remember:

$i^2=-1$

So the equation simplifies to:

$=cos(2\theta)+isin(2\theta)$

$=e^{i(2\theta)}$

**Plato** · March 24th 2011, 12:25 PM

Originally Posted by adam_leeds

is $(e^{i \theta})(e^{i \theta})$ = $e^{2i \theta})$

Yes it is.

BTW. See how I fixed the exponents is the LaTeX.
[tex] e^{2i \theta}[/tex] gives $e^{2i \theta}$ .
Note the braces about the exponents with more that one character.

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