## EarthCitizen 4 years ago z^4+z

1. anonymous

Factored: z(z^3+1)

2. EarthCitizen

how do you find the zeros ?

3. anonymous

oh, z^4 +1 = 0, z^4 = -1, anything to an even exponent is positive so there are no solutions

4. anonymous

In general you just factor the polynomial to find the 0's

5. anonymous

Oh, shoot, i did that wrong, its z^4 + z not + 1, sorry, let me refigure this

6. 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

7. TuringTest

...at least for the real solutions

8. EarthCitizen

yh, how is z^3=-1 plz ?

9. EarthCitizen

i get the z=0

10. TuringTest

(-1)(-1)(-1)=(1)(-1)=-1 so (-1)^3=-1

11. 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

12. EarthCitizen

yh, then shouldn't z=(-1)^1/3 ?

13. EarthCitizen

alryt, i see

14. TuringTest

is does...

15. TuringTest

(-1)^3=(-1)^(1/3)

16. EarthCitizen

does that still give us -1

17. 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

18. EarthCitizen

yh, tru

19. EarthCitizen

so the roots are z=0 and -1 ?

20. 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

21. EarthCitizen

no, plz what would the complex roots be ?

22. EarthCitizen

i mean do we use de moivre's theorem ?

23. 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

24. EarthCitizen

yes, so we should have four roots then ?

25. 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

26. TuringTest

fundamental theorem of algebra I mean*

27. EarthCitizen

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

28. EarthCitizen

th@ is 1<60 (polar form) is planar form it would be 1/2+j((3)^1/2)/2

29. EarthCitizen

$1/2+j \sqrt{3}/2$

30. 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...

31. EarthCitizen

alryt, L

32. EarthCitizen

does Euler's formula apply to the equation, z^4+z ?

33. 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

34. EarthCitizen

alryt, thanx

35. 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.