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

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

(b) Use deMoivre's Theorem to find the non-real solutions, z and z , of the equation z + 1 = 0.

Hence show that z =-z and z = -z .

Any help would be hugely appreciated !!

Cheers

*~* Pam xxx *~*

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# deMoivre's Theorem

Started by Pammy, May 10 2007 10:52 AM

2 replies to this topic

### #1

Posted 10 May 2007 - 10:52 AM

***~* 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

Posted 10 May 2007 - 08:03 PM

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

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

(b) Use deMoivre's Theorem to find the non-real solutions, z and z , of the equation z + 1 = 0.

Hence show that z =-z and z = -z .

Any help would be hugely appreciated !!

Cheers

*~* Pam xxx *~*

I could have sworn I replied to this thread earlier (a) Given that -1 = cos + i sin , - < , state the value of .

(b) Use deMoivre's Theorem to find the non-real solutions, z and z , of the equation z + 1 = 0.

Hence show that z =-z and z = -z .

Any help would be hugely appreciated !!

Cheers

*~* Pam xxx *~*

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

For (b), we have , so if you put (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 . Since the equation is a cubic, it has three roots. Letting n=0,1 and 2 in 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 . 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

Posted 11 May 2007 - 08:57 AM

Thanks sooooo much !!!

You have unfogged my wee brain !! hee hee !

*~* Pam xxx *~*

You have unfogged my wee brain !! hee hee !

*~* Pam xxx *~*

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