If $x=\frac{-1+i\sqrt{3}}{2}$ and $y=\frac{-1-i\sqrt{3}}{2}$, where $i^2=-1$, then which of the following is not correct?
$\mathrm{(A)\ } x^5+y^5=-1 \qquad \mathrm{(B) \ }x^7+y^7=-1 \qquad \mathrm{(C) \  } x^9+y^9=-1 \qquad$
$\mathrm{(D) \  } x^{11}+y^{11}=-1 \qquad \mathrm{(E) \  }x^{13}+y^{13}=-1$
Notice that $x+y=-1$ and $xy=1$. We have $1=(x+y)^2=x^2+2xy+y^2=x^2+y^2+2\implies x^2+y^2=-1$. 
We also have $x^3+y^3=(x+y)^3-3x^2y-3xy^2=(-1)^3-3xy(x+y)=-1-3(1)(-1)=2$.
Finally, $x^5+y^5=(x+y)^5-5x^4y-5xy^4-10x^3y^2-10x^2y^3$
$=(-1)^5-5xy(x^3+y^3+2xy(x+y))=-1-5(1)(2+2(1)(-1))=-1$.

Let $x^{2n+1}+y^{2n+1}=z$. We have 
$-z=(x^{2n+1}+y^{2n+1})(x^2+y^2)$
$-z=x^{2n+3}+y^{2n+3}+x^2y^{2n+1}+x^{2n+1}y^2$
$-z=x^{2n+3}+y^{2n+3}+(xy)^2(x^{2n-1}+y^{2n-1})$
$-z=x^{2n+3}+y^{2n+3}+x^{2n-1}+y^{2n-1}$.

For $n=2, z=-1$, so 
$1=x^7+y^7+x^3+y^3=x^7+y^7+2\implies x^7+y^7=-1$.

Therefore, for $n=3, z=-1$, and
$1=x^9+y^9+x^5+y^5=x^9+y^9-1\implies x^9+y^9=2$, so $x^9+y^9\not=-1, \boxed{\text{C}}$.

As a check, we have $z=2$ for $n=4$, and
$-2=x^{11}+y^{11}+x^7+y^7=x^{11}+y^{11}-1\implies x^{11}+y^{11}=-1$.

Finally, for $n=5$ we have $z=-1$, and
$1=x^{13}+y^{13}+x^9+y^9=x^{13}+y^{13}+2\implies x^{13}+y^{13}=-1$, and this is true for all other answer choices.