Complex roots

  • Thread starter msimard8
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  • #1
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Homework Statement



find all complex roots of z^8=81i

Homework Equations





The Attempt at a Solution



let the angle=x

z^8=r^8(cis8x)

we know

81i=81 (cis pi/2)

threfore

z^8=81(cos pi/2 + i sin (pi/2) )

8x= pi/2 + 2kpi
x = pi/16 + kpi/4 kEz

therefore

if k=1 z=sqrt3 (cis pi/16)

i go through this solutiosn...and should end up getting a result angle of pi/2 in one of them..becuse on of the roots is 1...so there i must be wrong..help
 

Answers and Replies

  • #2
Gib Z
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Why must one of the solutions be 1? 81i is not equal to 1.
 
  • #3
CompuChip
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Why do you think you can write z^8 = r^8 (cos 8x) (at least I assume that by cis you mean cos :smile:)?
You can certainly write it as r^8 e^{8 i x} for some angle x. Also, 81i can be written in such a way (you wrote it as 81(cos pi/2 + i sin pi/2) but I might as well write it as 81 e^(i pi/2)). Now r = (81)^(1/8). What equation(s) do you get for the angle x?
 
Last edited:
  • #4
Gib Z
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cis (x) is a notation engineers have adapted, an acronym for cos x + i sin x, which is equal to e^(ix).
 
  • #5
CompuChip
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Ah, I see. Strange people, those engineers :smile:
Anyway, your answer looks correct (at least, the one you gave, so I assume you have found the other 7 as well). Why did you think 1 was a solution?
 

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