2. $\alpha, \beta$ are the two imaginary roots of the quadratic equation $x^{2}+2 p x$ $+1=0$ with real coefficients, and the points corresponding to $\alpha, \beta, 1$ in the complex plane form an equilateral triangle. Then the real number $p=$ $\qquad$ .
2. $\frac{1}{2}$.

Since $\alpha, \beta$ are conjugate complex numbers, we have $|\alpha|=|\beta|$. And $\alpha \beta=1$, then $|\alpha \beta|=1$, which means $|\alpha|=|\beta|=1$. Therefore, the points corresponding to $\alpha, \beta, 1$ form an equilateral triangle inscribed in the unit circle, so
$$
\begin{array}{l}
\alpha=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}, \beta=-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}, \\
\alpha+\beta=-2 p=-1 . \\
\text { Hence } p=\frac{1}{2} .
\end{array}
$$