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Complex number question (1 Viewer)

cutemouse

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Hello,

I haven't come across this one before, except in this year's Ext 2 trial for my school. I have no idea on how to solve it. Could someone please help me out?

If ω is a non real solution to z6=1, show that ω64=-1

Thanks
 

azureus88

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interesting....but it doesnt seem rite cause w^6 + w^4 = -1 gives w^4 = -2 and LHS is not real. Unless w isnt complex root of unity with smallets argument...
 
D

DkAssain

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easy wuestions i try tomz
cant be bothered
looks interesting
 

gurmies

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I've made the same observation as azureus88, and am not sure how to interpret this anomalistic result. Seems incorrect to me..
 

cutemouse

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Hi guys,

I actually posted one part of the question by mistake, here is the whole question:

a) Find all the solutions to the equation z6=1 in the form of x+iy
b) If ω is a non real solution to z6=1, show that ω64=-1 [as above]
c) By choosing one particular value of ω, explain with the aid of a diagram why ω64=-1
 

cutemouse

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Ahh, I think I know how to do this now!!

Thanks everyone.
 

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