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    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...
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    z^5 = -1

    Look up the chapter on nth roots of unity.
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    I recall you do something along the lines of e^5(ni(pi)) = -1 or something, not sure, I should probably go and revise.
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    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?
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    ok thanks i got the answer,

    could a mod shut this thread down?
 
 
 
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