What roots of #z^4+z^2+1 = 0# satisfy #abs(z) < 1# ?

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Answer 1 · 2017-03-14T10:33:56

No root is in #abs z < 1#

Calling #u=z^2# we have

#u^2+u+1=0# solving for #u#
#u=1/2(-1pm isqrt3) = e^(pm i (phi + 2kpi))# with #phi=arctan sqrt(3)#

then

#z^2 = e^(pm i (phi+2kpi))# so #z=e^(pm i/2(phi+2kpi))#

but #abs (e^(ix))=1# so all four roots have unit modulus and no root is in #abs z < 1#

Answer 2 · 2017-05-21T18:15:55

None

Alternatively, note that:

#0 = (z^2-1)(z^4+z^2+1) = z^6-1#

So the roots are all #6#th roots of unity and therefore all satisfy #abs(z) = 1# and not #abs(z) < 1#.