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Complex roots and equations

  1. May 15, 2013 #1
    Question: http://gyazo.com/abca582e1109884964913493487ad8ae

    My solution:

    I got √6 + i√2 = √8(cos(pi/6) + isin(pi/6)) as they did below,
    then [itex] z^{\frac{3}{4}} = \sqrt{8}e^{i(\frac{\pi}{6} + 2k\pi)} [/tex]
    then took 4/3 of both sides and let k = 1, 0 etc to try and get the values of z

    however in the solutions: http://gyazo.com/3ba4f14d27f2fbc1bc8f04d2b918a5ac

    they raised it to 1/4 first then added the 2kpi and hence got different solutions than me. Could anyone explain why?
     
  2. jcsd
  3. May 15, 2013 #2

    Simon Bridge

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    The question is:

    Solve the equation ##z^{3/4}=\sqrt{6}+\sqrt{2}i##
    Put your answers in the form ##z=re^{i\theta}## where ##r>0## and ##-\pi < \theta \leq \pi##

    ... looking at the pic, it looks to me like they just raised both sides to the power of 4 to get the expression for ##z^3## ... if you take the cube root of both sides, it should be the same as your answer.
     
  4. May 15, 2013 #3

    haruspex

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    Your upload of the other solution is too hard to read. RHS is cropped. As far as I can tell, they also raised both sides to 4/3, effectively. Pls post your working and answer.
     
  5. May 16, 2013 #4
    What I done:

    [itex] z^{\frac{3}{4}} = \sqrt{8}e^{i(\frac{\pi}{6})} [/itex]
    as solutions are not unique:
    [itex] z^{\frac{3}{4}} = \sqrt{8}e^{i(\frac{\pi}{6} + 2k\pi)}[/itex]
    [itex] z = (\sqrt{8}e^{i(\frac{\pi}{6} + 2k\pi)})^{\frac{4}{3}} [/itex]

    now I subbed in k = 0, k = 1... to get the solutions where -π<θ≤pi

    what they have done
    [itex] z^{\frac{3}{4}} = \sqrt{8}e^{i(\frac{\pi}{6})} [/itex]
    [itex] z^3 = ( \sqrt{8}e^{i(\frac{\pi}{6})})^{4} [/itex]
    [itex] z^3 = 64e^{i(\frac{\pi}{6} + 2\pi k)} [/itex]
    so [itex] z = 4e^{i(\frac{2\pi}{9} + \frac{2k\pi}{3})} [/itex]

    then they sub in k = 0, k = 1 and k = -1 to get values for z (only one of which is the same to mine).
    I'm asking why do you have to add the 2pki after you get it into z^3 = ...
     
  6. May 16, 2013 #5

    haruspex

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    Ok, but what are your solutions? The last line above is still consistent with their solutions. Following on from it:
    [itex] z = 4(e^{i(\frac{2\pi}{9} + \frac{8k\pi}3)})[/itex]
    so the difference is a factor ##e^{i\frac{6k\pi}3}## = 1.
    I agree it would be more natural to introduce the k term straight away, but I'm not seeing that deferring until after multiplying by 4 makes a difference. The important thing is to do it before dividing.
     
  7. May 16, 2013 #6
    My only solution is [tex] z = 4e^{i(\frac{2\pi}{9})} [/tex], as all I can sub in is k = 0

    they have three solutions of:
    [tex] z = 4e^{i(\frac{2\pi}{9})} [/tex]
    [tex] z = 4e^{i(\frac{8\pi}{9})} [/tex]
    [tex] z = 4e^{i(\frac{-4\pi}{9})} [/tex]
     
  8. May 16, 2013 #7

    haruspex

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    Really? What do you get if you set k = 1?
     
  9. May 17, 2013 #8
    then you get theta as 26pi/9... but this is out of the domain given for theta.

    edit: hm, but as theta is not unique, is it OK to just take away 2pi from 26pi/9?
     
  10. May 17, 2013 #9

    haruspex

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    Exactly so. You should reduce all results mod 2pi to get them into the target range.
     
  11. May 18, 2013 #10
    thank you for your patience! :)
     
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