## Task B-4.6.

Determine all real numbers $x, y$ and $\alpha$ for which

$$
y=-2018 x \quad \text { and } \quad \frac{2018 x+i}{y+i}=\frac{1+i \sin \alpha}{1-i \sin 3 \alpha}, \quad x \neq 0, \alpha \in (0,2 \pi)
$$

$i$ denotes the imaginary unit $\left(i^{2}=-1\right)$.
## Solution.

After multiplying the second equality by the common denominator, we get

$$
\begin{gathered}
\frac{2018 x+i}{y+i}=\frac{1+i \sin \alpha}{1-i \sin 3 \alpha} \\
(y+i)(1+i \sin \alpha)=(2018 x+i)(1-i \sin 3 \alpha) \\
y-\sin \alpha+i(1+y \sin \alpha)=2018 x+\sin 3 \alpha+i(1-2018 x \sin 3 \alpha)
\end{gathered}
$$

By equating the imaginary parts of the left and right sides in (1) and substituting $y=-2018 x$, we get:

$$
\begin{gathered}
1+y \sin \alpha=1-2018 x \sin 3 \alpha \\
1-2018 x \sin \alpha=1-2018 x \sin 3 \alpha
\end{gathered}
$$

Since $x \neq 0$, it follows that $\sin \alpha=\sin 3 \alpha$.

The solutions to this equation are $\alpha=3 \alpha+2 k \pi$ or $\alpha=\pi-3 \alpha+2 k \pi$,

that is, $\alpha=k \pi$ or $\alpha=\frac{\pi}{4}+\frac{k \pi}{2}, k \in \mathbb{Z}$.

Since $\alpha \in (0, 2\pi)$, the solutions are $\alpha \in \left\{\frac{\pi}{4}, \frac{3 \pi}{4}, \pi, \frac{5 \pi}{4}, \frac{7 \pi}{4}\right\}$.

By equating the real parts in (1), we get

$$
\begin{aligned}
& y-\sin \alpha=2018 x+\sin 3 \alpha \\
& 2 y=\sin \alpha+\sin 3 \alpha \quad(\text{because } y=-2018 x \text{ and } \sin \alpha=\sin 3 \alpha) \quad 1 \text{ point} \\
& y=\sin \alpha
\end{aligned}
$$

If $\alpha=\pi$, then $y=x=0$, which is not a solution due to the condition $x \neq 0$ in the problem.

If $\alpha=\frac{\pi}{4}$ or $\frac{3 \pi}{4}$, then $y=\frac{\sqrt{2}}{2}, x=-\frac{1}{2018} \cdot \frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{4036}$.

If $\alpha=\frac{5 \pi}{4}$ or $\frac{7 \pi}{4}$, then $y=-\frac{\sqrt{2}}{2}, x=-\frac{1}{2018} \cdot \frac{-\sqrt{2}}{2}=\frac{\sqrt{2}}{4036}$.