1 Let $z$ be an imaginary number, and $z^{2}+\frac{9}{z^{2}}$ is a real number, then the value of $|z|$ is equal to
$1 \sqrt{3}$. If $\phi \neq 0$, then when and only when $r=1$, $r e^{-}+\frac{1}{r} e^{-\phi}=\left(r+\frac{1}{r}\right) \cos \phi + i\left(r-\frac{1}{r}\right) \sin \phi$ is a real number. Given that $\frac{z^{2}}{3}+\frac{3}{z^{2}}$ is a real number, so $\left|\frac{z^{2}}{3}\right|=1$, which means $|z|=\sqrt{3}$.