3. Given that the lengths of the three sides of $\triangle A B C$ are $a, b, c$, and the complex numbers $z_{1}, z_{2}$ satisfy: $\left|z_{1}\right|=a,\left|z_{2}\right|=b,\left|z_{1}+z_{2}\right|=$ $c$, then the complex number $\frac{z_{1}}{z_{2}}$ is (
(A) a real number
(B) a pure imaginary number
(C) not a real number
(D) not a pure imaginary number
3. (C).
$$
\begin{array}{l} 
\because \quad c^{2}=\left|z_{1}+z_{2}\right|=\left(z_{1}+z_{2}\right)\left(\overline{z_{1}}+\overline{z_{2}}\right) \\
=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-z_{1} \overline{z_{2}}+\overline{z_{1}} z_{2} \\
=a^{2}+b^{2}+\frac{z_{1}}{z_{2}} \cdot z_{2} \overline{z_{2}}+z_{1} \overline{z_{1}} \cdot \frac{z_{2}}{z_{1}} \\
=a^{2}+b^{2}+b^{2}\left(\frac{z_{1}}{z_{2}}\right)+a^{2}\left(\frac{z_{2}}{z_{1}}\right), \\
\therefore \quad b^{2}\left(\frac{z_{1}}{z_{2}}\right)^{2}+\left(a^{2}+b^{2}-c^{2}\right)\left(\frac{z_{1}}{z_{2}}\right)+a^{2}=0 .
\end{array}
$$

This is a real-coefficient linear equation in one variable about $\frac{z_{1}}{z_{2}}$, and $\Delta$
$$
=\left(a^{2}+b^{2}-c^{2}\right)-4 a^{2} b^{2}=(a+b+c)(a+b-c)(a-b
$$
$+c)(a-b-c)<0$, hence $\frac{z_{1}}{z_{2}}$ is an imaginary number (when and only when $a^{2}+b^{2}=$ $c^{2}$, i.e., $\triangle ABC$ is a right triangle, $\frac{z_{1}}{z_{2}}$ is a pure imaginary number).