complex exponential help please

**mewmew** · Oct3-05, 05:58 PM

Express the following in the form $LaTeX Code: z=Re[Ae^{i(\\omega t+\\alpha)}]$

$LaTeX Code: z=cos(\\omega t - \\frac{\\pi}{3}) - cos (\\omega t)$
and
$LaTeX Code: z=sin(\\omega t) - 2cos(\\omega t - \\frac{\\pi}{4}) + cos(\\omega t)$

I got a few of the problems correct by using trig. identities but it was pretty tough and two I can't get. Our teacher said you can use a tric to solve them easier but didn't have time to finish, I just know it has something to do with the polar form of $LaTeX Code: e^{i \\theta}$ I really have no clue on how to do these without using the really long method of trig. identities. Any help would be great. Thanks

**CarlB** · Oct4-05, 02:29 AM

$LaTeX Code: z = \\textrm{Re}(e^{i(\\omega t - \\pi/3} - e^{i\\omega t})$
$LaTeX Code: =\\textrm{Re}( e^{i\\omega t}( e^{-i\\pi/3}-1))$
$LaTeX Code: = \\textrm{Re}(( e^{-i\\pi/3}-1) \\;\\;e^{i\\omega t})$

Does that help?

Don't be ashamed, it's far better to be conversant in trig than to know a few tricks.

Carl

**robphy** · Oct4-05, 07:37 AM

Along the lines of CarlB, but without jumping straight in to using Re(),
recall $LaTeX Code: e^{i\\theta}=\\cos\\theta+i\\sin\\theta$ , from which you can derive
$LaTeX Code: \\cos\\theta=\\frac{1}{2}(e^{i\\theta}+e^{-i\\theta})$ and a similar expression for $LaTeX Code: \\sin\\theta$ (which I left for you to do).

So, now:
$LaTeX Code: <BR>z&=\\frac{1}{2}(e^{i[\\omega t-\\pi/3]}+e^{-i[\\omega t-\\pi/3]})-\\frac{1}{2}(e^{i[\\omega t]}+e^{-i[\\omega t]})$
then do some algebra.

Recall that $LaTeX Code: Re(z)=\\frac{1}{2}(z+z^*)$ . Thus $LaTeX Code: Re(e^{i\\theta})=\\cos\\theta$ .

**mewmew** · Oct4-05, 08:38 AM

Thanks a lot, that makes it much more simple.