1. The quadratic equation with complex coefficients $x^{2}+a x+b=0$ has exactly two pure imaginary roots, then ( )
(A) $a$ is zero, $b$ is a negative real number
(B) $a$ is zero, $b$ is a pure imaginary number
(C) $a$ is a pure imaginary number, $b$ is a real number
(D) $a$ is a pure imaginary number, $b$ is a pure imaginary number
,- 1. (C).
Let the roots of the equation be $\alpha i, \beta i$, where $\alpha, \beta$ are non-zero real numbers. By Vieta's formulas, we know that $\alpha i + \beta i = -a$, so $a = -(\alpha + \beta) i$, and $(\alpha i)(\beta i) = b$, so $b = -\alpha \beta$. Therefore, it can only be (C).