Suppose the roots to $z^4+az^3+bz^2+cz+d=0$ all have the property that $|z| =2$. Let $a,b,c,d \in \mathbb{C}$. Suppose the roots to $z^4+az^3+bz^2+cz+d=0$ all have the property that $|z| =2$. We want to prove that $\overline{a} = \frac{4c}{d}$.
I personally have no clue how to approach this problem. I have seen similar problems that use the same polynomial but ask different questions, such as whether the roots have negative real parts and so on. But for this on, I am completely lost on how to start it. Can we do something like
$$|z|^4+|a||z|^3+|b||z|^2+|c||z|+d =0$$
$$16+8|a|+4|b|+2c+d=0$$
But even here I am not sure how to proceed. Any advice on this?

Update: Each $z=2e^{i\theta}$ and so
$$16e^{i4\theta}+8ae^{i3\theta}+ 4be^{i2 \theta}+2ce^{i \theta}+d = 0$$
 A: suppose $z_1,z_2,z_3,z_4$ are the 4 roots.
$a=-(z_1+z_2+z_3+z_4)$
$\begin{align}
\overline{a} &= -(\overline{z_1}+\overline{z_2}+\overline{z_3}+\overline{z_4})\\
&=-(\frac{|z_1|^2}{z_1}+\frac{|z_2|^2}{z_2}+\frac{|z_3|^2}{z_3}+\frac{|z_4|^2}{z_4})\\
&=-4(\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}+\frac{1}{z_4})\\
&=-4\frac{z_1z_2z_3+z_2z_3z_4+\cdots}{z_1z_2z_3z_4}\\
&=\frac{4c}{d}
\end{align}$
A: Expanding on my comment, this is the original equation:
$$
z^4+az^3+bz^2+cz+d=0 \tag{1}
$$
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 $\,z^4\,$:
$$
256 + 64 \bar a z+16\bar b z^2 + 4 \bar c z^3 + \bar d z^4 = 0 \tag{2}
$$
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 $\,\lambda \ne 0\,$ such that $\,1 = \lambda \cdot \bar d\,$, $\,a = \lambda \cdot 4 \bar c\,$ etc. The first equality implies $\,d \ne 0\,$, 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.
