Let


* ${A\_j} \in {\mathbb{C}^{n \times n}},0<{w\_j}\in \mathbb{R} (j = 0,1,2....m)$ and $\lambda $ is a complex variable such that $\lambda=x+iy$ and $x,y\in \mathbb{R}$.
* ${\rm{P(}}\lambda {\rm{) = }}{{\rm{A}}\_m}{\lambda ^m} + .....{A\_1}\lambda + {A\_0}$ is a matrix polynomial.
* ${\rm{Q(}}\lambda {\rm{) = }}{{\rm{w}}\_m}{\lambda ^m} + .....{w\_1}\lambda + {w\_0}$
* t=$Q{(\left| \lambda \right|)^2}$


Why does $$D(x,y)=\det (tI - P{(\lambda )^\*}P(\lambda )) = \sqrt {{x^2} + {y^2}} p(x,y) + q(x,y)$$ where $p(x, y)$ and $q(x, y)$ are real polynomials in $x, y$? 


Furthermore, if $Q(x)$ is even function, then $p(x,y)=0 
$?



A question on determinant of a matrix polynomial
The statement is false if $P^\*$ is taken to mean the element by element complex conjugate of $P(\lambda)$. A counterexample: let $m=1$, $\omega\_0 = \omega\_1 = 1$, and
$$A\_0 = \left( \begin{array}{cc} 0&i\\2i&0 \end{array} \right) \\
A\_1 = \left( \begin{array}{cc} 1&i\\0&2 \end{array} \right)
$$
Then coefficients in $p(x,y)$ come out to be complex non-real.


You must have meant $P^\dagger(\lambda)P(\lambda)$, the Hermitian conjugate of the matrix.
With that change:


$t$ is a real polynomial in the variable $\lambda = \sqrt{x^2+y^2}$. 


Proposition 1:
$\forall n \in \Bbb{N} : \lambda^n $ is either a polynomial in $x^2$ and $y^2$ or (if $n$ is odd) $\sqrt{x^2+y^2}$ times a polynomial in $x^2$ and $y^2$.


Since all the $\omega\_m$ are real, we have by proposition 1 that $t = Q(|\lambda| = Q(\sqrt{x^2+y^2}$ is of the form 
$$
t = P\_1(x,y) + \sqrt{x^2+y^2}P\_2(x,y)
$$
where $P\_i(x,y)$ are both real polynomials.


Now for a given set of $A\_m$, each element of $P(\lambda)$ is a (possibly complex) polynomial in $(x,y)$. But each element of $P^\dagger(\lambda)P(\lambda)$ is a real-valued, thus it is a real polynomial in $(x,y)$.


Then each element of $tI - P^\dagger(\lambda)P(\lambda)$ is a real polynomial in $(x,y)$ plus, for diagonal elements, an expression of the form $P\_1(x,y) + \sqrt{x^2+y^2}P\_2(x,y)$. 


So each element of $tI - P^\dagger(\lambda)P(\lambda)$ is of the form $P\_1(x,y) + \sqrt{x^2+y^2}P\_2(x,y)$


Finally, the determinant of a matrix is a polnomial function of all of its elements. This brings us home, because any polynomial function of elements of the form $P\_1(x,y) + \sqrt{x^2+y^2}P\_2(x,y)$ is itself of the form 
$P\_3(x,y) + \sqrt{x^2+y^2}P\_4(x,y)$. Identify in your problem $q(x,y)$ with $P\_3$ and p(x,y) with $P\_4$.


By the way, if only even powers appear in $Q$, then $P\_2(x,y) = 0$ since every term ins itself a polynomial in $x^2+y^2)$. Since the off-diagonal elements are also pure polynomials in $x$ and $y$, in that situation, $p(x,y) = 0$.


