Question

z^4+z

Answer 1

Factored: z(z^3+1)

Answer 2

how do you find the zeros ?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

...at least for the real solutions

Answer 8

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

Answer 9

i get the z=0

Answer 10

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

Answer 11

in fact for all odd numbers n we have
(-x)^n=-x^n
and for all even n we have
(-x)^n=x^n

Answer 12

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

Answer 13

alryt, i see

Answer 14

is does...

Answer 15

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

Answer 16

does that still give us -1

Answer 17

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

Answer 18

yh, tru

Answer 19

so the roots are z=0 and -1 ?

Answer 20

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

Answer 21

no, plz what would the complex roots be ?

Answer 22

i mean do we use de moivre's theorem ?

Answer 23

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

Answer 24

yes, so we should have four roots then ?

Answer 25

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

Answer 26

fundamental theorem of algebra I mean*

Answer 27

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

Answer 28

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

Answer 29

\[1/2+j \sqrt{3}/2\]

Answer 30

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...

Answer 31

alryt, L

Answer 32

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

Answer 33

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

Answer 34

alryt, thanx

Answer 35

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.