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How to find all complex Z such that Z^5=-32

  • #1

Homework Statement


How to find all complex number Z such that Z^5=-32


Homework Equations


Euler equation
e^it=cost+isint


The Attempt at a Solution


I guess a naive way to solve is that since Z^5=(-2)^5
Therefore Z=-2, but this obviously too good to be true.
I have no background in complex analysis.
 

Answers and Replies

  • #2
Dick
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Z=(-2) isn't too good to be true. You are right. It is a root. But there are four more. Express -32 in polar form. I.e. -32=32*e^(it). The use deMoivre. What are the possibilities for t? There are more than one. Give me one to start out with.
 
Last edited:
  • #3
Thank you for your reply.
Sorry I do not understand that -32=32*e^(it)?
e^it=cost+isint
I mean t is a variable that takes on arbitrary values.
 
  • #4
Dick
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Thank you for your reply.
Sorry I do not understand that -32=32*e^(it)?
e^it=cost+isint
I mean t is a variable that takes on arbitrary values.
t isn't arbitrary if -32=32*e^(it). That means e^(it)=(-1). What's t?
 
  • #5
t=180degree=-5pi,-3pi,-pi,pi,3pi,5pi,7pi,9pi,11pi?
Therefore -32=32*e^(it)=32*[cos(pi)+isin[pi]]

Sorry how do I use de Moivre's formula? Never taken complex analysis
 
  • #6
Dick
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t=180degree=-5pi,-3pi,-pi,pi,3pi,5pi,7pi,9pi,11pi?
Therefore -32=32*e^(it)=32*[cos(pi)+isin[pi]]

Sorry how do I use de Moivre's formula? Never taken complex analysis
You could look it up rather than me telling you. You multiply the arguments (the angles) by 1/5. How many roots do you get before they start repeating?
 
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