$12 \cdot 24$ Consider the following equations:
( I ) $\sqrt{a^{2}+b^{2}}=0$;
( II ) $\sqrt{a^{2}+b^{2}}=a b$;
(III) $\sqrt{a^{2}+b^{2}}=a+b$;
(IV) $\sqrt{a^{2}+b^{2}}=a-b$.

Where $a, b$ represent real or complex numbers, and there are solutions other than $a=0$ and $b=0$ for the equation(s):
(A) ( I ), ( II ), ( III ), (IV).
(B) Only ( II ), (III), (IV).
(C) Only ( I ), (III),(IV).
(D) Only (III), (IV).
(E) Only ( I ).
(17th American High School Mathematics Examination, 1966)
[Solution] Note that for real numbers $a>1$, when $b=ai, \frac{a}{\sqrt{a^{2}}-1}, 0,0$, they respectively satisfy equations (I), (II), (III), (IV).
By the arbitrariness of $a$ $(a>1)$, the original equations all have infinitely many solutions.
Therefore, the answer is $(A)$.