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# FP3 Complex Numbers watch

1. find the solutions to z^5+1=0

i know that the the cube roots of 1 are (-1/2) (+/-) √3/2

but i'm just not sure what to do here...
2. z^5 = -1

Look up the chapter on nth roots of unity.
3. I recall you do something along the lines of e^5(ni(pi)) = -1 or something, not sure, I should probably go and revise.
4. i know that z = r(cosx + isinx)

therefore in this case it would be z^5 = r^5(cos5x+isin5x)

would i set r^5(cos5x+isin5x) = -1?
5. ok thanks i got the answer,

could a mod shut this thread down?

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Updated: March 22, 2009
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