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

Answer 1

Sophie Adams

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#

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Answer 2

Elias Ackerman

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#.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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