1. Given that $z$ is an imaginary number, and $z+\frac{9}{z}$ is a real number, then $|z|=$

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1. Given that $z$ is an imaginary number, and $z+\frac{9}{z}$ is a real number, then $|z|=$
$$
z+\frac{9}{z}=\bar{z}+\frac{9}{\bar{z}} \Rightarrow z-\bar{z}=\frac{9}{\bar{z}}-\frac{9}{z}=\frac{9(z-\bar{z})}{z \bar{z}}=\frac{9(z-\bar{z})}{|z|^{2}} \text {, }
$$

Since $z$ is an imaginary number $\Rightarrow z-\bar{z} \neq 0$, so $|z|^{2}=9 \Rightarrow|z|=3$.