[tex]\frac{z}{z^2-16} + \frac{2}{z+4}=\frac{z +2(z-4)}{(z -4)(z+4)} =\frac{z +2 z-8}{(z -4)(z+4)} =\frac{3z-8}{(z -4)(z+4)} \\ \\ \\ x^2-16 \neq 0 \\(x-4)(x+4)\neq 0 \\ x-4\neq 0 \ \ \ and \ \ \ x+4 \\ x \neq 4 \ \ \ and \ \ \ x \neq -4 \\D=R\setminus \left \{ -4,4 \right \}[/tex]
