n∈ℕ, p∈ℂ[x], ∀z∈ℂ* show $p(z+\frac{1}{z})=(z^n +\frac{1}{z^n})$ With $n∈ℕ$, Show that there exist a unique polynomial $p∈ℂ[x] $such that $∀z∈ℂ^*$, $p(z+1∕z)=(z^n +1/z^n)$.
 A: $(z + 1/z)^n = \displaystyle\Sigma_{k=0}^n \binom{n}{k}z^{n-k}z^{-k} = \Sigma_{k=0}^n \binom{n}{k}z^{n-2k}$
See where expanding the above gets you.
Detailed answer
First, consider the case $n$ even:
For $n = 0$ the polynomial is 2.
Suppose that it is possible to find a polynomial for any even number less than $n$.  Then expand
$$(z + 1/z)^n = z^n + nz^{n-2} + ... + nz^{-n + 2} + z^{-n}$$ (verify this expansion with the binomial expansion formula).  So now the problem is reduced to finding a polynomial in $z + 1/z$ that equals
$$-nz^{n-2} + ... - nz^{-n + 2} = -n(z^{n-2} + ... + z^{-n+2})$$ We know that a polynomial $P_{n-2}$ of $z + 1/z$ exists that equals $z^{n-2} + z^{-n+2}$, so the polynomial now has terms $w^n - nP_{n-2}(w)$.  We subsequently do the same thing over and over and, since the exponent of the polynomial to be found decreases by two, the algorithm eventually terminates at the base case and -- voilà -- we get our polynomial.
The case of $n$ odd is similar, only use n = 1 as the base case.
As far as uniqueness, suppose you had two different polynomials $P$ and $Q$ in $w$ that satisfied $P(z + 1/z) = Q(z + 1/z) = z^n + (1/z)^n$.  Then $P - Q$ would have degree at least one or be a constant.  The constant case could not stand, for the constant would have to be zero (meaning $P = Q$).  But if $P - Q$ had degree $n$ then we would have a polynomial with an uncountable number of zeros (i.e., the whole image of $z + 1/z$), which is a no-no by the Fundamental Theorem of Algebra.  So it must be that there is a unique polynomial satisfying the problem.
