TITLE: Which polynomial's roots are its coefficients?
QUESTION [41 upvotes]: Start with any polynomial of degree $n$ with complex coefficients, e.g.,
$$z^3+z^2+2 z+3 \;.$$
Find its $n$ roots, and list them in order of their modulus:
$$-1.28, (0.14\pm 1.53 i)$$
Now form a new polynomial with leading coefficient $1$ and the other
coefficients the $n$ roots:
$$
z^3
-1.28\, z^2
+(0.14 -1.53 i)\, z
+(0.14 +1.53 i) \;.
$$
Here I used the sorting convention that $(a-b i) < (a + b i)$.
A bit more formally, let $P$ be
$$
P \;:\; z^n + a_{n-1}z^{n-1} + \cdots + a_0 \;,
$$
with roots
$$r_{n-1},  \ldots ,r_0 \;,$$
sorted so that $|r_i| \le |r_{i-1}|$.
Define $P_1$ as
$$
P_1 \;:\; z^n + r_{n-1}z^{n-1} + \cdots + r_0 \;.
$$
My question is:

Q1. Which are the polynomials whose roots are its coefficients,
  in the sense that $r_i=a_i$, $i=0,\ldots,n-1$, i.e., $P_1 = P$?

Here is one fixed point:
$$
z^3
-(0.782599 +0.521714 i)\, z^2
+(0.884646 -0.589743 i)\, z
+(0.680552 +1.63317 i)
$$
When this process is iterated, it appears to fall into cycles, not always of length $1$.

Q2. Does iteration of this process always fall into a cycle?

(Added. 3Sep15.) The question suggested by Igor Rivin is also interesting,
especially in light of Richard Stanley's remark concerning real polynomials:

Q3. Which are the polynomials whose roots are its coefficients, in any order? (This is an unordered version of Q1.)

(Added. 4Sep15.)
Concerning Q2, which I realize is less interesting than the other questions, here is
an example of a cycle of length $11$. I plot the magnitude of the vector
of $3$ roots versus iteration #, starting from a random cubic polynomial. The last
vector of roots is
$$
(-1.52+0.88 i,-0.48-0.90 i,1.00 +0.02 i)
$$
with magnitude $2.27$:

REPLY [23 votes]: For the unordered version, the solutions for degree $3$ are
$$ \eqalign{x^3&\cr
x^3& {}+x^2-2\,x\cr
x^3&{}+x^2-x-1\cr
x^3&{}+ \left( r^2/2-1 \right) x^2+rx-r-r^2+2 \ \text{where}\ r^3 - 2 r + 2 = 0\cr}
$$
The solutions for degree $4$ are
$$ \eqalign{
x^4\cr
x^4&{}+x^3-x^2-x\cr
x^4&{}+x^3-2\; x^2 \cr
x^4&{}+ \left( \frac{r^2}{2}-1 \right) x^3+rx^2+(-r^2-r+2) x \ \text{where}\ r^3-2\,r+2 = 0\cr
x^4&{}-r^2 x+rx^2+x^3+r^2-2\,rx+r-x-1 \ \text{where}\
r^3+2\,r^2+r+1 = 0\cr
x^4&{}+ \left( \frac{r^{12}}{4}+{\frac {53\,r^{11}}{116}}+{\frac {19\,
r^{10}}{116}}-{\frac {43\,r^9}{116}}-{\frac {59\,r^8}{116}}-
{\frac {9\,{r}^{7}}{29}}+{\frac {11\,{r}^{6}}{29}}+{\frac {31\,{r}^{5}
}{29}}+{\frac {35\,{r}^{4}}{58}}\right.\cr &\left.{}-\frac{{r}^{3}}{29}+{\frac {2\,{r}^{2}}{29}
}+{\frac {5\,r^{13}}{116}}+\frac{r}{29}-{\frac {12}{29}} \right) {x}^{3}+r{x
}^{2}\cr &{}+ \left( -{\frac {49\,r^{11}}{29}}+{\frac {123\,{r}^{10}}{58}}+
{\frac {207\,{r}^{9}}{58}}+{\frac {195\,{r}^{8}}{58}}-{\frac {27\,{r}^
{7}}{58}}-{\frac {373\,{r}^{6}}{58}}-{\frac {113\,{r}^{5}}{29}}+{
\frac {77\,{r}^{4}}{29}}\right. \cr &\left.{}+{\frac {13\,{r}^{3}}{29}}-{\frac {47\,{r}^{13
}}{58}}-\frac{5\,{r}^{12}}{2}+{\frac {3\,{r}^{2}}{29}}-{\frac {100\,r}{29}}+{
\frac {98}{29}} \right) x+2\,{r}^{12}+{\frac {45\,{r}^{11}}{58}}-{
\frac {71\,{r}^{10}}{29}}\cr &{}-{\frac {82\,{r}^{9}}{29}}-{\frac {68\,{r}^{8
}}{29}}+{\frac {63\,{r}^{7}}{58}}+{\frac {329\,{r}^{6}}{58}}+{\frac {
51\,{r}^{5}}{29}}-{\frac {112\,{r}^{4}}{29}}+{\frac {21\,{r}^{13}}{29}
}-{\frac {74}{29}}-{\frac {11\,{r}^{3}}{29}}\cr&{}-{\frac {7\,{r}^{2}}{29}}+
{\frac {69\,r}{29}}\ \text{where}\cr
 {r}^{14}&{}+2\,{r}^{13}-{r}^{12}-4\,{r}^{11}-{r}^{10}+4\,{r}^{8}+6\,{r}^{
7}-4\,{r}^{6}-6\,{r}^{5}+4\,{r}^{4}+4\,{r}^{2}-8\,r+4
=0
}$$
Note that you can always multiply a solution by $x^k$ for positive integer $k$, obtaining $k$ more coefficients of $0$ to match $k$ more roots of $0$.