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

##### 2 Answers
Mar 14, 2017

No root is in $\left\mid z \right\mid < 1$

#### Explanation:

Calling $u = {z}^{2}$ we have

${u}^{2} + u + 1 = 0$ solving for $u$
$u = \frac{1}{2} \left(- 1 \pm i \sqrt{3}\right) = {e}^{\pm i \left(\phi + 2 k \pi\right)}$ with $\phi = \arctan \sqrt{3}$

then

${z}^{2} = {e}^{\pm i \left(\phi + 2 k \pi\right)}$ so $z = {e}^{\pm \frac{i}{2} \left(\phi + 2 k \pi\right)}$

but $\left\mid {e}^{i x} \right\mid = 1$ so all four roots have unit modulus and no root is in $\left\mid z \right\mid < 1$

May 21, 2017

None

#### Explanation:

Alternatively, note that:

$0 = \left({z}^{2} - 1\right) \left({z}^{4} + {z}^{2} + 1\right) = {z}^{6} - 1$

So the roots are all $6$th roots of unity and therefore all satisfy $\left\mid z \right\mid = 1$ and not $\left\mid z \right\mid < 1$.