Let $a$ and $b$ be real numbers so that the roots of
\[z^2 + (10 + ai) z + (27 + bi) = 0\]are complex conjugates.  Enter the ordered pair $(a,b).$
I know that complex conjugates are of the form $c \pm di$, where $c$ and $d$ are real numbers. Simplifying, I get that $2c = - 10 - ai$ and $c^2 + d^2 = 27 + bi$. Substituting $c = -5$ into the second equation, I get that $25 + d^2 = 27 + bi$. This implies that $d^2 = 2 + bi$. Since $d$ is a real number, I need $bi = 0$. Therefore, $b = 0$ and $d = \pm \sqrt{2}$. So, the ordered pair $(a,b)$ is $(0,0)$. 