Suppose the roots to z 4 + a z 3 + b z 2 + c z + d = 0 all have the property that | z | = 2

Question

Suppose the roots to       z    4    +  a      z    3    +  b      z    2    +  c  z  +  d  =  0 all have the property that       |    z      |    =  2

Nola Rivera · Accepted Answer

suppose       z    1    ,      z    2    ,      z    3    ,      z    4   are the 4 roots.  a  =  −  (      z    1    +      z    2    +      z    3    +      z    4    )                              a          ¯                                    =        −        (                              z            1                    ¯                +                              z            2                    ¯                +                              z            3                    ¯                +                              z            4                    ¯                )                                          =        −        (                                            |                                      z              1                                                      |                            2                                            z            1                          +                                            |                                      z              2                                                      |                            2                                            z            2                          +                                            |                                      z              3                                                      |                            2                                            z            3                          +                                            |                                      z              4                                                      |                            2                                            z            4                          )                                          =        −        4        (                  1                      z            1                          +                  1                      z            2                          +                  1                      z            3                          +                  1                      z            4                          )                                          =        −        4                                            z              1                                      z              2                                      z              3                        +                          z              2                                      z              3                                      z              4                        +            ⋯                                              z              1                                      z              2                                      z              3                                      z              4                                                                        =                              4            c                    d

polivijuye · Accepted Answer

Expanding on my comment, this is the original equation:

\begin{matrix} (1) & z^{4} + a z^{3} + b z^{2} + c z + d = 0 \end{matrix}

Taking the conjugate:

{\bar{z}}^{4} + \bar{a} {\bar{z}}^{3} + \bar{b} {\bar{z}}^{2} + \bar{c} \bar{z} + \bar{d} = 0

All roots have magnitude 2 so

\bar{z} = \frac{| z |^{2}}{z} = \frac{4}{z}

, then after substituting and multiplying by

z^{4}

\begin{matrix} (2) & 256 + 64 \bar{a} z + 16 \bar{b} z^{2} + 4 \bar{c} z^{3} + \bar{d} z^{4} = 0 \end{matrix}

Equations (1) and (2) have the same roots, and therefore they must be identical up to a multiplicative factor, so the coefficients must be proportional:

(1, a, b, c, d) \propto (\bar{d}, 4 \bar{c}, 16 \bar{b}, 64 \bar{a}, 256)

This means there exists a

λ \neq 0

such that

1 = λ \cdot \bar{d}

etc. The first equality implies

d \neq 0

, then dividing the second equality by the first one and taking the conjugates on both sides gives the relation

\bar{a} = \frac{4 c}{d}

that OP's question asks for.

Suppose the roots to z 4 </msup> + a z 3 </msup> + b z

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