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deMoivre's Theorem - HSN forum

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deMoivre's Theorem


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#1 Pammy

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Posted 10 May 2007 - 10:52 AM

Got really stuck at these questions and was wondering if anyone could maybe explain them to me ??

(a) Given that -1 = cos theta.gif + i sin theta.gif , -pi.gif < theta.gif lessorequal.gif pi.gif, state the value of theta.gif.

(b) Use deMoivre's Theorem to find the non-real solutions, z base1.gif and z base2.gif, of the equation z power3.gif + 1 = 0.
Hence show that z base1.gif power2.gif=-z base2.gif and z base2.gif power2.gif= -z base1.gif.


Any help would be hugely appreciated !!
Cheers
*~* Pam xxx *~*
*~* Kiss me, beneath the milky twilight, lead me, out on the moonlit floor, lift your open hand, strike up the band and make the fireflies dance, silver moon sparkling, so kiss me *~*

#2 ad absurdum

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Posted 10 May 2007 - 08:03 PM

QUOTE(Pammy @ May 10 2007, 11:52 AM) View Post
Got really stuck at these questions and was wondering if anyone could maybe explain them to me ??

(a) Given that -1 = cos theta.gif + i sin theta.gif , -pi.gif < theta.gif lessorequal.gif pi.gif, state the value of theta.gif.

(b) Use deMoivre's Theorem to find the non-real solutions, z base1.gif and z base2.gif, of the equation z power3.gif + 1 = 0.
Hence show that z base1.gif power2.gif=-z base2.gif and z base2.gif power2.gif= -z base1.gif.


Any help would be hugely appreciated !!
Cheers
*~* Pam xxx *~*
I could have sworn I replied to this thread earlier dry.gif

For (a), you need to note that we can write -1 as z=-1 + 0i. If z=\cos\theta + i\sin\theta, we can say \cos\theta = -1 and \sin\theta = 0. Letting theta be pi gives this.

For (b), we have z^3=-1, so if you put z^3 = \cos(\pi + 2n\pi) + i\sin(\pi + 2n\pi) (what was found in the first part, except also note that you get the same value if we are an even number of pi's further on). Use DeMoivre to find z, so z = [\cos(\pi + 2n\pi) + i\sin(\pi + 2n\pi)]^{\frac{1}{3}} = [\cos(\frac{\pi + 2n\pi}{3}) + i\sin(\frac{\pi + 2n\pi}{3})]. Since the equation z^3=-1 is a cubic, it has three roots. Letting n=0,1 and 2 in z=[\cos(\frac{\pi + 2n\pi}{3}) + i\sin(\frac{\pi + 2n\pi}{3})] will give you these roots (two should be complex and one real, the real one being the obvious one).
Let your two complex roots be z_1, z_2. For the next part you need to square these respectively and compare the value to the other root. You can square them using DeMoivre's theorem, or just by multiplying them out.
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#3 Pammy

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Posted 11 May 2007 - 08:57 AM

Thanks sooooo much !!!
You have unfogged my wee brain !! hee hee !
*~* Pam xxx *~*
*~* Kiss me, beneath the milky twilight, lead me, out on the moonlit floor, lift your open hand, strike up the band and make the fireflies dance, silver moon sparkling, so kiss me *~*





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