Calculus/roots

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Expert: - 2/25/2009


Question
express answers in a+bi..

x^6 + 729 = 0


I did x= sqrt(-729)
     x= i sqrt(729)
     x= 3i

(3i is not one of the answers when k=0)

So what I did was...

3, when k=0
3(cos(2pi/6)+isin(cos(2p/6))) when k=1
3cos(4pi/6)+isin(cos(4p/6)))   when k=2
3(cos(6pi/6)+isin(cos(6p/6)))   when k=3
3(cos(8pi/6)+isin(cos(8p/6)))    when k=4
3(cos(10pi/6)+isin(cos(10p/6)))  when k=5

i got x as 3i...

but just using the angles i got..

k=1
3/2 + 3sqrt3/2 i
k=2
-3/2 + 3sqrt3/2 i
k=3
-1
k=4
-3/2 - 3sqrt3/2 i
k=5
3/2-3sqrt3/2 i




I cannot figure out what I'm doing wrong, I've tried it many times and still the answers is wrong..  

Answer
We have x^6 + 729 = 0, or x^6 = -729.

Note that the answer is already at π degrees in polar coordinates,
since it is negative,
and that's where the major mistake can be found.

The polar number is 3(cos(π/6)+sin(π/6)).

The corrected answers are
3(cos( 1π/6) + i*sin( 1π/6)) when k=0.
3(cos( 3π/6) + i*sin( 3p/6)) when k=1
3(cos( 5π/6) + i*sin( 5p/6)) when k=2
3(cos( 7π/6) + i*sin( 7p/6)) when k=3
3(cos( 9π/6) + i*sin( 9p/6)) when k=4
3(cos(11π/6) + i*sin(11p/6)) when k=5

I think that answers your question.

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