What can one conclude for a poynomial with the property that p(x)=p(ix)?

Question

From a theorem in the theory of cyclotomic polynomials I deduced that a special polynomial $p(x)$ of even degree $n$ has the property $$p(x)=p(ix)$$ with $i$ being the complex unit.
What can one conclude from this ? I am especialy interested in calculating the Taylor expansion of the polynomial. What I see already from CAS calculation is that only even powers of $x$ appear in the polynomial and I hope this may be inferred from the property somehow.
But if I develop the Taylor series around $a=0$ I can "prove" that all coefficients are zero...
Maybe there is some more tricky center $a$ to do the development around ?

Answer 1

If $p(x) = p(ix)$, then $p^{(k)}(0) = i^k p^{(k)}(0)$ from which we see that $p^{(k)}(0) = 0$ unless $4 \mid k$. Hence $p$ is of the form $p(x) = \sum_k p_k x^{4k}$.

If $p$ has the form $p(x) = \sum_k p_k x^{4k}$, we see that $p(x) = p(ix)$, hence the set of such polynomials is $\operatorname{sp} \{ x \mapsto x^{4k}\}_{k=0}^\infty$.

Answer 2

If $p(ix)=p(x)$ then $p(-x)=p(x)$ as well, that is $p$ is even and can be written as $p(x)=q(x^2)$ for a suitable polynomial $q$. Then $q(-x)=q(x)$ as well, i.e. $q$ is even so that ultimately $p(x)=r(x^4)$ for a polynomial $r$. Conversely, for any polynomial $r$, we can define $p(x)=r(x^4)$ and have $p(ix)=r((ix)^4)=r(x^4)=p(x)$ as desired.