What roots of z4+z2+1=0 satisfy |z|&lt;1?

Step 1 Calling u=z2 we have u2+u+1=0 solving for u u=12(−1±i3)=e±i(ϕ+2kπ) with ϕ=arctan⁡3 then z2=e±i(ϕ+2kπ) so z=e±i2(ϕ+2kπ) but |eix|=1 so all four roots have unit modulus and no root is in |z|&lt;1

Get the answer to "What roots of z4+z2+1=0 satisfy |z|<1?"

Aubrey Hendricks

Answered question

2022-01-30

What roots of

z^{4} + z^{2} + 1 = 0

satisfy

| z | < 1

Answer & Explanation

Amari Larsen

Beginner2022-01-31Added 10 answers

Step 1
Calling

u = z^{2}

we have

u^{2} + u + 1 = 0

solving for u

u = \frac{1}{2} (- 1 \pm i \sqrt{3}) = e^{\pm i (ϕ + 2 k π)}

with

ϕ = \arctan \sqrt{3}

then

z^{2} = e^{\pm i (ϕ + 2 k π)}

z = e^{\pm \frac{i}{2} (ϕ + 2 k π)}

but

| e^{i x} | = 1

so all four roots have unit modulus and no root is in

| z | < 1

Kingston Gates

Beginner2022-02-01Added 8 answers

Step 1
Alternatively, note that:

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

So the roots are all 6th roots of unity and therefore all satisfy

| z | = 1

and not

| z | < 1

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