## Task B-4.4.

Let the complex number $z=(a+\cos \theta)+(\sqrt{3} a-\sin \theta) i$. Determine for which real numbers $a$ it holds that $|z| \leq 3$, for all $\theta \in \mathbb{R}$.
## Solution.

$$
|z|^{2}=(a+\cos \theta)^{2}+(\sqrt{3} a-\sin \theta)^{2}
$$

After rearrangement, we get

$$
|z|^{2}=4 a^{2}+1+2 a(\cos \theta-\sqrt{3} \sin \theta)=4 a^{2}+1+4 a\left(\frac{1}{2} \cos \theta-\frac{\sqrt{3}}{2} \sin \theta\right)
$$

or

$$
|z|^{2}=4 a^{2}+1+4 a \sin \left(\frac{\pi}{6}-\theta\right)
$$

From the condition $|z| \leq 3$, we have

$$
4 a^{2}+1+4 a \sin \left(\frac{\pi}{6}-\theta\right) \leq 9
$$

or

$$
a \sin \left(\frac{\pi}{6}-\theta\right) \leq 2-a^{2}
$$

One possibility is $a=0$. (1)

For $a>0$, we have $\sin \left(\frac{\pi}{6}-\theta\right) \leq \frac{2-a^{2}}{a}$. This will hold for all $\theta \in \mathbb{R}$ if $\frac{2-a^{2}}{a} \geq-1$, or if $a^{2}-a-2 \leq 0$. We conclude $a \in\langle 0,2]$

For $a<0$, we have $\sin \left(\frac{\pi}{6}-\theta\right) \geq \frac{2-a^{2}}{a}$. This will hold for all $\theta \in \mathbb{R}$ if $\frac{2-a^{2}}{a} \leq 1$, or if $a^{2}+a-2 \leq 0$. We conclude $a \in[-2,0\rangle$. (3)

From (1), (2), and (3), we get $a \in[-2,2]$.