Question 105: If for all $\theta \in R$, the modulus of the complex number $z=(a+\cos \theta)+(2 a-\sin \theta)$ i does not exceed 2, then the range of the real number $a$ is $\qquad$ _.
Question 105, Solution: The condition is equivalent to: for any $\theta \in \mathbb{R}$, we have
$$
\begin{array}{l}
(a+\cos \theta)^{2}+(2 a-\sin \theta)^{2} \leq 4 \\
\Leftrightarrow 5 a^{2}+2(\cos \theta-2 \sin \theta) a-3 \leq 0
\end{array}
$$

The range of $\cos \theta-2 \sin \theta$ is $[-\sqrt{5}, \sqrt{5}]$, so (*) is equivalent to:
$$
5 a^{2}+2 \sqrt{5}|a|-3 \leq 0 \Leftrightarrow|a| \leq \frac{\sqrt{5}}{5}
$$

Therefore, the range of $a$ is $\left[-\frac{\sqrt{5}}{5}, \frac{\sqrt{5}}{5}\right]$.