$12 \cdot 23$ If the 6 roots of the equation $x^{6}=-64$ are all written in the form $a+b i$, where $a, b$ are real numbers, then the product of the roots with real part $a>0$ is 
(A) -2.
(B) 0.
(C) $2 i$.
(D) 4.
(E) 16.
(41st American High School Mathematics Examination, 1990)
[Solution] Since $64=2^{6}$, the given equation can be transformed into $x^{6}=-2^{6}$, thus its 6 roots are all distributed on a circle with a radius of 2 and centered at point $O$ (note that one of the roots is $2i$).
Obviously, for the complex roots $z_{5}, z_{6}$, their real parts are greater than 0.
$$
\text { Also, } \quad \begin{aligned}
z_{5} & =2\left[\cos \left(-\frac{\pi}{6}\right)+i \sin \left(-\frac{\pi}{6}\right)\right] \\
& =2\left(\frac{\sqrt{3}}{2}-\frac{1}{2} i\right)=\sqrt{3}-i,
\end{aligned}
$$

and $z_{6}=\sqrt{3}+i$,
then $z_{5} \cdot z_{6}=(\sqrt{3}-i)(\sqrt{3}+i)=4$.
Therefore, the answer is $(D)$.