7. Let $a$ be a real number. The equation $\left(z^{2}-2 z+5\right)\left(z^{2}+2 a z+1\right)=0$ has 4 distinct roots, and the 4 points corresponding to these roots in the complex plane are concyclic. Then the range of the real number $a$ is $\qquad$

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If $\Delta=4\left(a^{2}-1\right)0$, then the 4 roots correspond to 4 points in the complex plane: $(1,2),(1,-2),\left(-a-\sqrt{a^{2}-1}, 0\right),\left(-a+\sqrt{a^{2}-1}, 0\right)$. Thus, the center of the circle is $(-a, 0)$, leading to $\sqrt{(a+1)^{2}+4}=\sqrt{a^{2}-1} \Rightarrow a=-3$. Therefore, the range of the real number $a$ is $(-1,1) \cup\{-3\}$.