$4 \cdot 4$ Given the following two propositions:
(1) Let $a, b, c$ all be complex numbers. If $a^{2}+b^{2}>c^{2}$, then $a^{2}+b^{2}-c^{2}>0$;
(2) Let $a, b, c$ all be complex numbers. If $a^{2}+b^{2}-c^{2}>0$, then $a^{2}+b^{2}>c^{2}$. Then, which of the following statements is correct?
(A) Proposition (1) is correct, and proposition (2) is also correct.
(B) Proposition (1) is correct, and proposition (2) is incorrect.
(C) Proposition (1) is incorrect, and proposition (2) is also incorrect.
(D) Proposition (1) is incorrect, and proposition (2) is correct.
(China High School Mathematics League, 1994)
［Solution］According to the problem, we have:
Proposition (1) is correct. $a^{2}+b^{2}>c^{2}$ indicates that $a^{2}+b^{2}$ and $c^{2}$ are real numbers. Therefore, according to the rule of transposition, we have $a^{2}+b^{2}-c^{2}>0$.

Proposition (2) is incorrect. $a^{2}+b^{2}-c^{2}>0$ only indicates that $a^{2}+b^{2}-c^{2}$ is a real number, but it does not guarantee that $a^{2}+b^{2}$ and $c^{2}$ are both real numbers, so $a^{2}+b^{2}>c^{2}$ does not necessarily hold.
For example, take $a=2+i, b=i, c=\sqrt{2}+\sqrt{2} i$, then we have
$$
a^{2}+b^{2}-c^{2}=(3+4 i)+(-1)-4 i=2>0,
$$

but it is not true that $a^{2}+b^{2}=2+4 i>4 i=c^{2}$.
Therefore, the answer is (B).