3. Let $p, q$ be complex numbers $(q \neq 0)$, if the roots of the equation $x^{2}+p x+q^{2}=0$ have equal modulus, then the value of $\frac{p}{q}$ is ( ).
A. Pure imaginary number
B. Non-zero complex number
C. Real number
D. Cannot be determined
3. C

Let the two roots be $x_{1}, x_{2}$. Then $\left\{\begin{array}{l}x_{1}+x_{2}=-p \\ x_{1} \cdot x_{2}=q^{2}\end{array}\right.$, considering $\frac{p^{2}}{q^{2}}=\frac{\left(x_{1}+x_{2}\right)^{2}}{x_{1} x_{2}}=\frac{x_{2}}{x_{1}}+\frac{x_{1}}{x_{2}}+2$ and since $\left|Z_{1}\right|=\left|Z_{2}\right| \Rightarrow Z_{1} \cdot \bar{Z}_{1}=Z_{2} \cdot \bar{Z}_{2} \Rightarrow \frac{Z_{1}}{Z_{2}}=\left(\frac{\bar{Z}_{2}}{Z_{1}}\right)$. Therefore, $\frac{p^{2}}{q^{2}}=\left(\frac{x_{1}}{x_{2}}\right)+\frac{x_{1}}{x_{2}}+2=2 \operatorname{Re}\left(\frac{x_{1}}{x_{2}}\right)+2 \geq 0$, so $\frac{p}{q} \in \mathbf{R}$.