I already posted this question at MSE [here](https://math.stackexchange.com/questions/685767/minimize-norm-of-a-polynomial-on-a-circle), but as it received no significative feedback for a while I cross-post it here.


I also noticed a related question [here](https://mathoverflow.net/questions/35317/minimizing-the-modulus-of-a-polynomial-around-a-circle) on MO (which does not answer my present question).


Let $P=\sum\_{k=0}^n a\_kX^k$ ba a polynomial of degree $n \gt 0$, and let
$r\gt 0$. Suppose that $P$ is not the monomial $a\_nX^n$, in other words
there is at least an $i<n$ such that $a\_i\neq 0$. Denote by $C$ the 
circle $\big\lbrace z \ \big| \ |z|=r\big\rbrace$. Clearly,
the continuous function $|P|$ attains a minimum value (let us denote it
by $\mu$) on the compact set $C$. Let $M=\big\lbrace z\in C
\ \big| \ |P(z)|=\mu\big\rbrace$.
 How many elements can we have in $M$ ? When $\mu=0$, $M$ contains
 only roots of $P$, so that the maximum cardinality is $n$. 


For $l\in[-n,n]$, let us put 
 $$b\_l=\sum\_{j=-n+{\sf max}(0,-l)}^{n+{\sf min}(0,-l)}
 \bar{a\_j}a\_{j+l}r^{2j+l}, B(X)=\sum\_{l=-n}^{n}b\_lX^{l+n}, D(X)=\frac{B(X)}{X^n}
 $$.


Then we have the identity $|P(re^{i\theta})|^2=D(e^{i\theta})$ for any $\theta\in{\mathbb R}$.
 Since $D'(X)=\frac{XB'(X)-B(X)}{X^{n+1}}$, we see that 
 $|M| \leq |M'|$ where $M'=\big\lbrace z \in {\mathbb C} \ \big| \ zB’(z)-B(z)=0, |z|=1
 \big\rbrace$. Remark that $D'$ is zero iff $B$ is a monomial in $X$, iff
 $P$ itself is a monomial in $X$. As the degree of $XB'(X)-B(X)$ is exactly $2n$, we deduce $|M| \leq 2n$.


Note however that we have only counted local extrema here, and the question is about the global extrema. Thus, one expects $|M|$ to be significatively
 lower than the upper bound $2n$. In fact, I conjecture the following :


**Conjecture.** $|M| \leq n$. 


I have checked this conjecture on a few random numerical examples.
 Does anyone have an idea about how to prove or find a counterexample to
 this conjecture ?



Minimize norm of a polynomial around a circle (count the solutions)
The conjecture is correct. 


Let me begin with a simple proof that $|M|\leq 2n$.


The curve $|P(z)|^2=\mu^2$ is a real algebraic curve of degree $2n$.
At the points of $M$ this curve must be tangent to the unit circle. The unit circle is
a curve of degree $2$, therefore by Bezout theorem these two curves have at most $4n$
intersections counting multiplicity. But the multiplicity is at least $2$ at each point. 


Now suppose WLOG that $1$ is not in $M$.
Consider the rational function 
$$f(z)=P\left(\frac{1}{1-z}\right).$$ 
The preimage of the unit circle under $1/(1-z)$ is
a straight line $L$. And the set $|f(z)|^2=\mu^2$ is an algebraic curve $C$ of degree $2n$.
The intersection $L\cap C$ may consist of at most $2n$ points, counting multiplicity.
But the multiplicity of each point is at least $2$. This proves the statement.


