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EarthCitizen

  • 4 years ago

z^4+z

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  1. anonymous
    • 4 years ago
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    Factored: z(z^3+1)

  2. EarthCitizen
    • 4 years ago
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    how do you find the zeros ?

  3. anonymous
    • 4 years ago
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    oh, z^4 +1 = 0, z^4 = -1, anything to an even exponent is positive so there are no solutions

  4. anonymous
    • 4 years ago
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    In general you just factor the polynomial to find the 0's

  5. anonymous
    • 4 years ago
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    Oh, shoot, i did that wrong, its z^4 + z not + 1, sorry, let me refigure this

  6. TuringTest
    • 4 years ago
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    z^4+z=0 z(z^3+1)=0 so by the zero factor rule we have z=0 and/or z^3+1=0 z^3=-1 z=-1 so z={-1,0} where the multiplicity of -1 is 3 and the multiplicity of 0 is 1

  7. TuringTest
    • 4 years ago
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    ...at least for the real solutions

  8. EarthCitizen
    • 4 years ago
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    yh, how is z^3=-1 plz ?

  9. EarthCitizen
    • 4 years ago
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    i get the z=0

  10. TuringTest
    • 4 years ago
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    (-1)(-1)(-1)=(1)(-1)=-1 so (-1)^3=-1

  11. TuringTest
    • 4 years ago
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    in fact for all odd numbers n we have (-x)^n=-x^n and for all even n we have (-x)^n=x^n

  12. EarthCitizen
    • 4 years ago
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    yh, then shouldn't z=(-1)^1/3 ?

  13. EarthCitizen
    • 4 years ago
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    alryt, i see

  14. TuringTest
    • 4 years ago
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    is does...

  15. TuringTest
    • 4 years ago
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    (-1)^3=(-1)^(1/3)

  16. EarthCitizen
    • 4 years ago
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    does that still give us -1

  17. TuringTest
    • 4 years ago
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    sure x=(-1)^(1/3) asks what number x is such that (x)(x)(x)=-1 the only real answer is -1 because (-1)(-1)(-1)=-1 so (-1)^(1/3)=-1

  18. EarthCitizen
    • 4 years ago
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    yh, tru

  19. EarthCitizen
    • 4 years ago
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    so the roots are z=0 and -1 ?

  20. TuringTest
    • 4 years ago
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    the real roots, yes there are also the complex solutions like -(-1)^(2/3) but I omitted those because they are rarely used in high-school level problems

  21. EarthCitizen
    • 4 years ago
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    no, plz what would the complex roots be ?

  22. EarthCitizen
    • 4 years ago
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    i mean do we use de moivre's theorem ?

  23. TuringTest
    • 4 years ago
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    that and (-1)^(1/3) but I know that will prompt you to ask me 'but you just said that (-1)^(1/3)=-1' but there is a complication that I glossed over there where we should use De Moivere's and stuff, yes

  24. EarthCitizen
    • 4 years ago
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    yes, so we should have four roots then ?

  25. TuringTest
    • 4 years ago
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    all real solutions z={-1,0} complex solutions z={-(-1)^(2/3),(-1)^(1/3)} a total of four by the Fundamental theorem of calculus you can check that the complex solutions work because they satisfy z^3=-1

  26. TuringTest
    • 4 years ago
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    fundamental theorem of algebra I mean*

  27. EarthCitizen
    • 4 years ago
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    when i tried solving for the imaginary roots i got 1<60 how did you come about -(-1)^2/3..

  28. EarthCitizen
    • 4 years ago
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    th@ is 1<60 (polar form) is planar form it would be 1/2+j((3)^1/2)/2

  29. EarthCitizen
    • 4 years ago
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    \[1/2+j \sqrt{3}/2\]

  30. TuringTest
    • 4 years ago
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    I'm not so good at doing these analytically this one is just easy to remember because (-1)^(1/3) is something you have to look at anyway, and -(-1)^(2/3)=-1^(1/3) so they have a nice pattern aside from that kind of thing I have to look it up, I've not done much complex analysis here's a page with some info on roots of http://tutorial.math.lamar.edu/Extras/ComplexPrimer/Roots.aspx perhaps you have done more of that kind of thing than I have, but I can try to use De Moive, hold on...

  31. EarthCitizen
    • 4 years ago
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    alryt, L

  32. EarthCitizen
    • 4 years ago
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    does Euler's formula apply to the equation, z^4+z ?

  33. TuringTest
    • 4 years ago
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    I'm not getting the right answer from de-moivre, but wolfram says it's http://www.wolframalpha.com/input/?i=solve+z%5E4%2Bz%3D0 yes, euler's formula is used in de moivre

  34. EarthCitizen
    • 4 years ago
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    alryt, thanx

  35. TuringTest
    • 4 years ago
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    yeah, this is trickier than I thought analytically de moivre doesn't seem to cover it directly, because it's not the power or root of a complex number. and because it's negative it's not a root of unity, so I'm not sure what form of analysis will give you the answer. Sure it has to do with euler's formula though.

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