z^4+z

- EarthCitizen

z^4+z

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

Factored: z(z^3+1)

- EarthCitizen

how do you find the zeros ?

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

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

- 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

...at least for the real solutions

- EarthCitizen

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

- EarthCitizen

i get the z=0

- TuringTest

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

- 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

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

- EarthCitizen

alryt, i see

- TuringTest

is does...

- TuringTest

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

- EarthCitizen

does that still give us -1

- 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

yh, tru

- EarthCitizen

so the roots are z=0 and -1 ?

- 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

no, plz what would the complex roots be ?

- EarthCitizen

i mean do we use de moivre's theorem ?

- 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

yes, so we should have four roots then ?

- 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

fundamental theorem of algebra I mean*

- EarthCitizen

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

- EarthCitizen

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

- EarthCitizen

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

- 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

alryt, L

- EarthCitizen

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

- 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

alryt, thanx

- 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.