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Z^4=-80i or general z^n=a+bi

  1. Feb 20, 2017 #1
    1. The problem statement, all variables and given/known data

    I just cant seem to get the right answer. z^4+80i=0
    looking at the complex plane u see the radius=r=80 (obviously)

    using De Moivre extension: z^n=(r^(1/n))(cos((x/n)+k2pi/n)-isin((x/n)+k2pi/n)


    z1=((80)^(1/4))(cos(3pi/8)+isin(3pi/8)

    shouldnt this be a root?


    z2= -((80)^(1/4))(cos(3pi/8)+isin(3pi/8)

    z3=((80)^(1/4))(cos((3pi/8)+(k2pi/8)))+isin((3pi/8)+(k2pi/8))

    for k=1, 2.

    what am i doing wrong?

    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Feb 20, 2017 #2
    Shouldnt it be k = { 0 , 1, 2, 3} I used this formula to find roots: (edited)

    [tex] 80*e^{-i\frac{((3/4)\Pi) + 2\Pi*K }{4}} [/tex]
     
    Last edited: Feb 20, 2017
  4. Feb 20, 2017 #3
    yes but first root already has k=0
     
  5. Feb 20, 2017 #4

    FactChecker

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    Yes. This looks ok to me. Is there some reason you think it is not?
    This also looks correct to me. Maybe I'm overlooking something. What makes you think that you are wrong?
    For the homework format, you should separate your work into the appropriate sections, not just append empty sections.
     
  6. Feb 20, 2017 #5
    getting wrong answer from the machine...
     
  7. Feb 20, 2017 #6

    FactChecker

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    This looks wrong to me. 'i' should multiply the entire argument and the argument of -80i is 3π/2 or -π/2.
     
  8. Feb 20, 2017 #7
    yes, sorry. [tex] 80*e^{-i\frac{((3/2)\Pi) + 2\Pi*K }{4}} [/tex]
     
  9. Feb 20, 2017 #8
    so in short my answer is correct?
     
  10. Feb 20, 2017 #9

    FactChecker

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    I think so. Maybe something went wrong when you checked it.
     
  11. Feb 20, 2017 #10

    SammyS

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    There are at least a couple of errors here.

    Your z1 looks good.

    z2 is not. How can you have a negative modulus?

    Write ## -80i\ ## in terms of a negative angle, and work with that, (if that's what you intended by the negative).

    For z3 (& z4 ?): Why are you dividing 2π by 8 ? It's not an 8th root that you want, is it?
     
  12. Feb 20, 2017 #11
    because when z^n and n is an even number the first roots are +-

    if z1 is good then i dont know what the problem is. as for z3 z4 i probably wrote them wrong. point is that the argument is (3pi/2)/n in this case n=4

    so 3pi/8 + k2pi/8

    edit: yes its its a mistake it shouldb k2pi/4. thanks. strange that i got a wrong answer for z1 though.
     
  13. Feb 20, 2017 #12

    FactChecker

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    It's just -z1, which is another 4'th root.
    Good catch. That is a mistake.
     
  14. Feb 20, 2017 #13

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    But shouldn't it be 3pi/8 + k2pi/4 ?
     
  15. Feb 20, 2017 #14
    yes realized that now thanks. but i didnt use any k value for z1 and i still got the wrong answer =(.
     
  16. Feb 20, 2017 #15

    FactChecker

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    You should re-check your check with the computer calculation. I'm suspicious of that.
     
  17. Feb 20, 2017 #16
    edit: i feel like the biggest idiot on the planet. its 81 not 80....

    thanks to everyone for their help-
     
  18. Feb 20, 2017 #17

    FactChecker

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    No problem. The error that @SammyS spotted was important to find anyway. If you didn't make the 80 versus 81 mistake, that probably would have never been found. :>)
     
  19. Feb 20, 2017 #18
    hello OP can i sugges you use latex editor online, I am learning it but this is much quicker and your work looks cleaner (so easy to spot any errors):

    [tex] z^{4}= -80i [/tex]https://www.codecogs.com/latex/eqneditor.php
     
  20. Feb 21, 2017 #19
    thanks will check it out.
     
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