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Factored: z(z^3+1)

how do you find the zeros ?

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

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

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

...at least for the real solutions

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

i get the z=0

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

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

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

alryt, i see

is does...

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

does that still give us -1

yh, tru

so the roots are z=0 and -1 ?

no, plz what would the complex roots be ?

i mean do we use de moivre's theorem ?

yes, so we should have four roots then ?

fundamental theorem of algebra I mean*

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

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

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

alryt, L

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

alryt, thanx