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EarthCitizen
 4 years ago
z^4+z
EarthCitizen
 4 years ago
z^4+z

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EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.1how do you find the zeros ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh, z^4 +1 = 0, z^4 = 1, anything to an even exponent is positive so there are no solutions

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In general you just factor the polynomial to find the 0's

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, shoot, i did that wrong, its z^4 + z not + 1, sorry, let me refigure this

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0z^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
 4 years ago
Best ResponseYou've already chosen the best response.0...at least for the real solutions

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.1yh, how is z^3=1 plz ?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0(1)(1)(1)=(1)(1)=1 so (1)^3=1

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0in fact for all odd numbers n we have (x)^n=x^n and for all even n we have (x)^n=x^n

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.1yh, then shouldn't z=(1)^1/3 ?

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.1does that still give us 1

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0sure 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
 4 years ago
Best ResponseYou've already chosen the best response.1so the roots are z=0 and 1 ?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0the real roots, yes there are also the complex solutions like (1)^(2/3) but I omitted those because they are rarely used in highschool level problems

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.1no, plz what would the complex roots be ?

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.1i mean do we use de moivre's theorem ?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0that 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
 4 years ago
Best ResponseYou've already chosen the best response.1yes, so we should have four roots then ?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0all 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
 4 years ago
Best ResponseYou've already chosen the best response.0fundamental theorem of algebra I mean*

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.1when i tried solving for the imaginary roots i got 1<60 how did you come about (1)^2/3..

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.1th@ is 1<60 (polar form) is planar form it would be 1/2+j((3)^1/2)/2

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.1\[1/2+j \sqrt{3}/2\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0I'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
 4 years ago
Best ResponseYou've already chosen the best response.1does Euler's formula apply to the equation, z^4+z ?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0I'm not getting the right answer from demoivre, 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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, 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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