EarthCitizen
  • EarthCitizen
z^4+z
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Factored: z(z^3+1)
EarthCitizen
  • EarthCitizen
how do you find the zeros ?
anonymous
  • anonymous
oh, z^4 +1 = 0, z^4 = -1, anything to an even exponent is positive so there are no solutions

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anonymous
  • anonymous
In general you just factor the polynomial to find the 0's
anonymous
  • anonymous
Oh, shoot, i did that wrong, its z^4 + z not + 1, sorry, let me refigure this
TuringTest
  • TuringTest
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
TuringTest
  • TuringTest
...at least for the real solutions
EarthCitizen
  • EarthCitizen
yh, how is z^3=-1 plz ?
EarthCitizen
  • EarthCitizen
i get the z=0
TuringTest
  • TuringTest
(-1)(-1)(-1)=(1)(-1)=-1 so (-1)^3=-1
TuringTest
  • TuringTest
in fact for all odd numbers n we have (-x)^n=-x^n and for all even n we have (-x)^n=x^n
EarthCitizen
  • EarthCitizen
yh, then shouldn't z=(-1)^1/3 ?
EarthCitizen
  • EarthCitizen
alryt, i see
TuringTest
  • TuringTest
is does...
TuringTest
  • TuringTest
(-1)^3=(-1)^(1/3)
EarthCitizen
  • EarthCitizen
does that still give us -1
TuringTest
  • TuringTest
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
EarthCitizen
  • EarthCitizen
yh, tru
EarthCitizen
  • EarthCitizen
so the roots are z=0 and -1 ?
TuringTest
  • TuringTest
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
EarthCitizen
  • EarthCitizen
no, plz what would the complex roots be ?
EarthCitizen
  • EarthCitizen
i mean do we use de moivre's theorem ?
TuringTest
  • TuringTest
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
EarthCitizen
  • EarthCitizen
yes, so we should have four roots then ?
TuringTest
  • TuringTest
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
TuringTest
  • TuringTest
fundamental theorem of algebra I mean*
EarthCitizen
  • EarthCitizen
when i tried solving for the imaginary roots i got 1<60 how did you come about -(-1)^2/3..
EarthCitizen
  • EarthCitizen
th@ is 1<60 (polar form) is planar form it would be 1/2+j((3)^1/2)/2
EarthCitizen
  • EarthCitizen
\[1/2+j \sqrt{3}/2\]
TuringTest
  • TuringTest
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...
EarthCitizen
  • EarthCitizen
alryt, L
EarthCitizen
  • EarthCitizen
does Euler's formula apply to the equation, z^4+z ?
TuringTest
  • TuringTest
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
EarthCitizen
  • EarthCitizen
alryt, thanx
TuringTest
  • TuringTest
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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