9. Given $z \in \mathbf{C}$. If the equation in $x$
$$
4 x^{2}-8 z x+4 i+3=0
$$

has real roots. Then the minimum value of $|z|$ is $\qquad$
9. 1 .

Let $z=a+b \mathrm{i}(a, b \in \mathbf{R}), x=x_{0}$ be the real root of the given equation. Then
$$
\begin{array}{l}
4 x_{0}^{2}-8(a+b \mathrm{i}) x_{0}+4 \mathrm{i}+3=0 \\
\Rightarrow\left\{\begin{array}{l}
4 x_{0}^{2}-8 a x_{0}+3=0, \\
-8 b x_{0}+4=0 .
\end{array}\right.
\end{array}
$$

Eliminating $x_{0}$ and rearranging gives
$$
\begin{array}{l}
3 b^{2}-4 a b+1=0 \Rightarrow a=\frac{3 b^{2}+1}{4 b} . \\
\text { Hence }|z|^{2}=a^{2}+b^{2}=\left(\frac{3 b^{2}+1}{4 b}\right)^{2}+b^{2} \\
=\frac{25}{16} b^{2}+\frac{1}{16 b^{2}}+\frac{3}{8} \geqslant \frac{5}{8}+\frac{3}{8}=1,
\end{array}
$$

with equality holding if and only if $b= \pm \frac{\sqrt{5}}{5}$.