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Linear algebra Homework question

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data
    I am entirely lost with this one question I can't seem to figure out how to do it at all. The question states that [tex]\omega[/tex] is a complex number where [tex]\omega[/tex]=(r)(e^(i [tex]\theta[/tex]))
    r and [tex]\theta[/tex] are real numbers
    r>0
    [tex]\theta[/tex] is element of [0,2[tex]\pi[/tex][
    n is a positive integer

    consider the equation z^n = [tex]\omega[/tex]
    Solve for z in terms of r, [tex]\theta[/tex] and n

    2. Relevant equations

    I would love to list some relevant equations but like I said, I have absolutely no idea what to do and any help at all would be greatly appreciated.


    3. The attempt at a solution

    I have tried looking over it multiple times however nothing seems to click. If anyone can just simply put me in the right direction or link me to anything that might help, that would be great!


    Thank you for your help,
    -Nick
     
  2. jcsd
  3. Nov 1, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Nickcanada! Welcome to PF! :smile:

    (have a theta: θ and a pi: π and an omega: ω and try using the X2 tag just above the Reply box :wink:)

    You're looking for a solution in the form z = se, such that zn = re :wink:
     
  4. Nov 1, 2009 #3
    Hey :) Thanks for the welcome as well as the help.

    I'm not too sure where to find all those symbols really :( New too all this!

    and I guess that I replace z in the equation with z=se
    so that Zn= [se]n = re

    but then from there, how do you isolate each variable or constant so that you get some sort of answer?
    sorry :( I'm just having a really hard time with this one...

    Thanks again!
     
  5. Nov 1, 2009 #4

    tiny-tim

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    ok, and when you expand it, [se]n = … ? :smile:
     
  6. Nov 1, 2009 #5
    well it ends up being

    Zn=sn eiφn= re

    but the path still seems blocked to me.... :(
     
  7. Nov 2, 2009 #6

    tiny-tim

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    (just got up :zzz: …)

    ok, sneinφ= re

    so now use the fact that the polar form, re, is unique in r, and unique in θ (mod 2π) :wink:
     
  8. Nov 4, 2009 #7
    Sir Tiny-tim is my solution to the problem right?

    zn = rei[tex]\theta[/tex]
    ln(zn) = ln(rei[tex]\theta[/tex])
    nlnz = ln(rei[tex]\theta[/tex])
    lnz = [tex]1/n[/tex]ln(rei[tex]\theta[/tex])
    lnz = ln(rei[tex]\theta[/tex])1/[tex]n[/tex]
    z = (rei[tex]\theta[/tex])1/n
     
  9. Nov 5, 2009 #8

    tiny-tim

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    (just got up :zzz: …)
    erm :redface: … it's very long, and it isn't a solution.

    You've started with zn = rei[tex]\theta[/tex], and ended with z = (rei[tex]\theta[/tex])1/n,

    which is just a restatement of the question.

    As I said, the solution has to be in the form z = se

    you have to say what s is, and what φ is, in each solution. :smile:
     
  10. Nov 5, 2009 #9
    But the problem is to solve for z, so I was thinking i just need to have z in one side and other parameters in the other side(Am I right???).
     
  11. Nov 5, 2009 #10

    tiny-tim

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    Yes, the LHS of your z = (re)1/n is excellent.

    It's the RHS that isn't finished. :wink:
     
  12. Nov 7, 2009 #11
    Hey! Sorry been busy for the past couple of days :(
    Still stuck on this question though..
    I'm slowly understanding how it works though :)

    Soo Sneinφ=re
    I'm supposed to solve for z, but does that mean that I have to solve for s as well as φ individually or not?... which are the modulus and the argument I believe?

    and so z= (sneinφ)1/n how do I add in the
    θ [tex]\in[/tex] [0,2π[ I know it has to go in there somewhere!!

    Thanks again,
    -Nick
     
  13. Nov 8, 2009 #12

    tiny-tim

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    Hey Nick! :smile:

    (just got up :zzz: …)
    Yes, that's correct :smile:

    if sneinφ=re,

    then that's exactly the same as saying sn = r and einφ = e
    (have an ε :wink:)

    Yes! :biggrin:

    If einφ = e,

    then that's exactly the same as saying that nφ = θ (mod 2π), or nφ = θ + a multiple of 2π. :smile:
     
  14. Nov 8, 2009 #13
    ok okay thanks :) soo then

    do I replace the φ with θ+2kπ? or rather nφ

    and that's it? Or is there more to be done?

    and shouldn't the answer in polar form be with φ as opposed to θ?

    Thanks again :)
     
  15. Nov 8, 2009 #14

    tiny-tim

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    erm :redface:make up your mind!

    φ = … ? :smile:
    Not following you. :confused:
     
  16. Nov 8, 2009 #15
    I guess it's nφ = θ+2k so I substitute nφ for that.

    My question is.. is that where it ends?

    if the question states: put the solutions in the form z=se would we keep nφ or still substitute it for θ+2k?

    I'm getting myself confused :S

    Thanks!!

    -Nick
     
  17. Nov 8, 2009 #16

    tiny-tim

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    Hi Nick! :smile:

    (btw, you mean 2π, not 2)

    Look, it's very simple …

    your answer has to start with "φ =" on the left-hand side …

    you haven't done that yet!

    φ = … ?​
     
  18. Nov 8, 2009 #17
    Sorry I copy pasted it but apparently the pi didn't work :S

    Okay so then the answer for what φ equals to is

    φ= (θ+2kπ)/n

    and sn=r so

    s= r1/n

    and that gives me the argument and the modulus
    so now to solve for z I write

    z= r1/nei(θ+2kπ)/n
    = (rei(θ+2kπ))1/n

    Thanks again btw!! :)
    -Nick
     
  19. Nov 9, 2009 #18

    HallsofIvy

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    Stop here!

     
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