A complex number problem of equation ${z}^4 + {z}^3 + {2z}^2 + {z} + 1 = 0$

Question

If $z$ be a complex number satisfying ${z}^4 + {z}^3 + {2z}^2 + {z} + 1 = 0$, then find the value of $|\bar{z}|$

How to you proceed with this problem?

Replacing $z$ with $a + {i}b$ doesn't seem to work, as you just get a huge biquadratic equation. I do recognize that $|\bar{z}| = |z|$, but not sure how to manipulate this fact to simplify the given biquadratic equation

See https://math.stackexchange.com/questions/480102/quadratic-substitution-question-applying-substitution-p-x-frac1x-to-2x4x — lab bhattacharjee, Nov 30 '17 at 09:04

José Carlos Santos · Accepted Answer · 2017-11-30T16:48:11.993

2

Let $w=z+\frac1z$. Then\begin{align}z^4+z^3+2z^2+z+1=0&\iff z^2+z+2+\frac1z+\frac1{z^2}=0\\&\iff\left(z+\frac1z\right)^2+z+\frac1z=0\\&\iff w^2+w=0\\&\iff w=0\vee w=-1.\end{align}So, solve the equations$$z+\frac1z=0\text{ and }z+\frac1z=-1.$$You'll see that, if $z$ is a solution, then you always get $|z|=1$. So $|\overline z|=|z|=1$.

edited Nov 30 '17 at 16:48

answered Nov 30 '17 at 09:06

José Carlos Santos

427,504

How have you identified the substitution? – lab bhattacharjee Nov 30 '17 at 09:23
@labbhattacharjee This is a standard substitution for a palindrome equation i.e. one that reads the same from front to back, as is the case here. – Sarvesh Ravichandran Iyer Nov 30 '17 at 09:25
@labbhattacharjee Actually, I saw first that we would get a very symmetric equation observing that$$z^4+z^3+2z^2+z+1=0\iff z^2+z+2+\frac1z+\frac1{z^2}=0$$and that gave me the idea of working with $z+\frac1z$. – José Carlos Santos Nov 30 '17 at 09:26
@JoséCarlosSantos, This is formally called Reciprocal Equation See https://math.stackexchange.com/questions/480102/quadratic-substitution-question-applying-substitution-p-x-frac1x-to-2x4x – lab bhattacharjee Nov 30 '17 at 09:50
@labbhattacharjee Thank you. I didn't know that. – José Carlos Santos Nov 30 '17 at 09:50

score 1 · Answer 2 · answered Nov 16 '20 at 21:22

1

Note $${z}^4 + {z}^3 + {2z}^2 + {z} + 1 =( z^4 + z^3 + z^2 )+(z^2 + z+ 1 ) \\= (z^2+1)(z^2 + z+ 1 ) =\frac{ (z^2+1)(z^3-1 ) }{z-1}=0 $$ which leads to $z^2=-1$, or $z^3=1$. As a result, $|\bar z|=1$.

answered Nov 16 '20 at 21:22

Quanto

97,352

A complex number problem of equation ${z}^4 + {z}^3 + {2z}^2 + {z} + 1 = 0$

2 Answers2