7. Given $z \in \mathbf{C}$. If the equation $x^{2}-2 z x+\frac{3}{4}+\mathrm{i}=0$ (where $\mathrm{i}$ is the imaginary unit) has real roots, then the minimum value of $|z|$ is . $\qquad$
7.1.

Let $z=a+b \mathrm{i}(a, b \in \mathbf{R}), x=x_{0}$ be a real root of the original equation.
$$
\begin{array}{l}
\text { Then } x_{0}^{2}-2\left(a+b \text { i) } x_{0}+\frac{3}{4}+\mathrm{i}=0\right. \\
\Rightarrow\left\{\begin{array}{l}
x_{0}^{2}-2 a x_{0}+\frac{3}{4}=0, \\
-2 b x_{0}+1=0
\end{array}\right. \\
\Rightarrow x_{0}=\frac{1}{2 b}, a=\frac{3 b^{2}+1}{4 b} \\
\Rightarrow|z|^{2}=a^{2}+b^{2}=\left(\frac{3 b^{2}+1}{4 b}\right)^{2}+b^{2} \\
\quad=\frac{25}{16} b^{2}+\frac{1}{16 b^{2}}+\frac{3}{8} \\
\geqslant \frac{5}{8}+\frac{3}{8}=1,
\end{array}
$$

When and only when $b= \pm \frac{\sqrt{5}}{5}$, the equality holds.
Thus, the minimum value of $|z|$ is 1.