IMC 2024 · Day 1 · P1

Hint: Write $a = e^{ix}$ and $b = e^{iy}$, and transform the RHS to a product.

Write $a = e^{ix}$ and $b = e^{iy}$ for some $x, y \in [0, 2\pi)$. Using Euler's formula, and the well-known identities $$\sin x + \sin y = 2 \sin\frac{x+y}{2} \cos\frac{x-y}{2} \quad \text{and} \quad \sin x = 2 \sin\frac{x}{2} \cos\frac{x}{2},$$ we get a product form of the left-hand side as $$\begin{align*} \operatorname{Im} \bigl( a + b + a\bar{b} \bigr) &= \sin x + \sin y + \sin(x - y) \\ &= 2 \sin\frac{x+y}{2} \cos\frac{x-y}{2} + 2 \sin\frac{x-y}{2} \cos\frac{x-y}{2} \\ &= 2 \left( \sin\frac{x+y}{2} + \sin\frac{x-y}{2} \right) \cos\frac{x-y}{2} \\ &= 4 \sin\frac{x}{2} \cdot \cos\frac{y}{2} \cdot \cos\frac{x-y}{2}. \end{align*}$$ Hence, $a + b + a\bar{b}$ is real if and only if either $\sin\frac{x}{2} = 0$, $\cos\frac{y}{2} = 0$ or $\cos\frac{x-y}{2} = 0$, which respectivelly correspond to $x = 2k\pi$, $y = (2k+1)\pi$ and $x = y + (2k+1)\pi$.

Therefore, the solutions are $$(1, b), \quad (a, -1) \quad \text{and} \quad (a, -a) \quad \text{with } |a| = 1, \ |b| = 1.$$

Notice that $$a + b + a\bar{b} \in \mathbb{R} \iff 1 + a + b + a\bar{b} \in \mathbb{R}.$$ Let $c \in \mathbb{C}$ be such that $a = c^2$. Now observe that $$\begin{align*} \bar{c} (1 + a + b + a\bar{b}) &= \bar{c} + \bar{c} c^2 + \bar{c} b + \bar{c} c^2 \bar{b} \\ &= \bar{c} + c + \bar{c} b + c \bar{b} \in \mathbb{R}, \end{align*}$$ where we used that $c \bar{c} = 1$ and $z + \bar{z} \in \mathbb{R}$ for any $z \in \mathbb{C}$. We conclude that either $c \in \mathbb{R}$, or $1 + a + b + a\bar{b} = 0$. In the first case, $c = \pm 1$ and so $a = 1$. In the second case, we factor the equation as $$(a + b)(1 + \bar{b}) = 1 + a + b + a\bar{b} = 0, \quad \text{and as such, } a = -b \text{ or } b = -1.$$ We find precisely three families of pairs $(a, b)$: the pairs $(1, b)$ for $b$ on the unit circle; the pairs $(a, -1)$ for $a$ on the unit circle; and the pairs $(a, -a)$ for $a$ on the unit circle.