2. Determine the pairs of complex numbers $(u, v)$ with $|u|=|v|=1$ such that

$$
|1-u|+\left|v^{2}+1\right|=|1-v|+\left|u^{2}+1\right|=\sqrt{2}
$$

Marius Damian, Brăila
Algebraic solution. Let $u=a+bi$ with $a^2+b^2=1$ and $v=c+di$ with $c^2+d^2=1$. Then $|1-u|+|v^2+1|=\sqrt{2} \Leftrightarrow |1-a-bi|+|c^2+2cdi-d^2+1|=\sqrt{2} \Leftrightarrow \sqrt{(1-a)^2+b^2}+|2c^2+2cdi|=\sqrt{2} \Leftrightarrow$ $\Leftrightarrow \sqrt{1-2a+a^2+b^2}+2|c|\cdot|c+di|=\sqrt{2} \Leftrightarrow \sqrt{2-2a}+2|c|=\sqrt{2}$, so $2-2a \geq 0 \Leftrightarrow a \leq 1$ and $\sqrt{2-2a} \leq \sqrt{2} \Leftrightarrow 2-2a \leq 2 \Leftrightarrow a \geq 0$, so $a \in [0,1]$. Similarly, $\sqrt{2-2c}+2|a|=\sqrt{2} \Rightarrow c \in [0,1]$.

$$
\left\{\begin{array}{l}
\sqrt{2-2a}+2c=\sqrt{2} \\
\sqrt{2-2c}+2a=\sqrt{2}
\end{array} \Rightarrow \sqrt{2-2a}+2c=\sqrt{2-2c}+2a \Leftrightarrow \sqrt{2-2a}-\sqrt{2-2c}=2a-2c \Leftrightarrow\right.
$$

$$
\Leftrightarrow \frac{2-2a-2+2c}{\sqrt{2-2a}+\sqrt{2-2c}}=2a-2c \Rightarrow a=c
$$

Thus, $\sqrt{2-2a}+2a=\sqrt{2} \Rightarrow \sqrt{2-2a}=\sqrt{2}-2a$ with $\sqrt{2}-2a \geq 0 \Rightarrow a \leq \frac{\sqrt{2}}{2}$ and $2-2a=2-4\sqrt{2}a+4a^2 \Rightarrow 4a^2=4\sqrt{2}a-2a$.
I. $a=0$;

II. $a \neq 0 \Rightarrow 4a=4\sqrt{2}-2 \Leftrightarrow a=\sqrt{2}-\frac{1}{2}$. But $\sqrt{2}-\frac{1}{2} \leq \frac{\sqrt{2}}{2} \Rightarrow 2\sqrt{2}-1 \leq \sqrt{2} \Rightarrow \sqrt{2} \leq 1$, which is false!

Therefore, $a=c=0, b= \pm 1, d= \pm 1$.

The problem has four solutions: $(u, v) \in \{(i, i), (-i, i), (i, -i), (-i, -i)\}$.

Trigonometric solution. Write $u, v$ in trigonometric form. There exist $\alpha, \beta \in [0, 2\pi)$ such that $u=\cos \alpha + i \sin \alpha$ and $v=\cos \beta + i \sin \beta$.

Then $\frac{\alpha}{2} \in [0, \pi) \Rightarrow \sin \frac{\alpha}{2} \in [0, 1]$ and $\frac{\beta}{2} \in [0, \pi) \Rightarrow \sin \frac{\beta}{2} \in [0, 1]$, so

$$
|1-u|=|(1-\cos \alpha) - i \sin \alpha| = \sqrt{2-2\cos \alpha} = \sqrt{4 \sin^2 \frac{\alpha}{2}} = 2 \sin \frac{\alpha}{2}
$$

and similarly $|1-v|=2 \sin \frac{\beta}{2}$.

Also,

$$
\begin{aligned}
& \left|u^2+1\right|=|(1+\cos 2\alpha) + i \sin 2\alpha| = \sqrt{2+2\cos 2\alpha} = \sqrt{4 \cos^2 \alpha} = 2|\cos \alpha| = 2\left|1-2 \sin^2 \frac{\alpha}{2}\right| \text{ and similarly } \\
& \left|v^2+1\right|=2\left|1-2 \sin^2 \frac{\beta}{2}\right|
\end{aligned}
$$

Therefore, using the hypothesis, we have:

$$
2\sqrt{2} = |1-u| + |1-v| + \left|u^2+1\right| + \left|v^2+1\right| = \left(2 \sin \frac{\alpha}{2} + 2\left|1-2 \sin^2 \frac{\alpha}{2}\right|\right) + \left(2 \sin \frac{\beta}{2} + 2\left|1-2 \sin^2 \frac{\beta}{2}\right|\right)
$$

But

$$
2 \sin \frac{\alpha}{2} + 2\left|1-2 \sin^2 \frac{\alpha}{2}\right| = \left\{\begin{array}{c}
2 \sin \frac{\alpha}{2} + 2 - 4 \sin^2 \frac{\alpha}{2}, \text{ if } \frac{\alpha}{2} \in \left[0, \frac{\pi}{4}\right] \cup \left[\frac{3\pi}{4}, \pi\right] \\
2 \sin \frac{\alpha}{2} + 4 \sin^2 \frac{\alpha}{2} - 2, \text{ if } \frac{\alpha}{2} \in \left(\frac{\pi}{4}, \frac{3\pi}{4}\right)
\end{array}\right.
$$

so the cases appear:

- If $\frac{\alpha}{2} \in \left[0, \frac{\pi}{4}\right] \cup \left[\frac{3\pi}{4}, \pi\right]$, then

$$
E(\alpha) = 2 \sin \frac{\alpha}{2} + 2\left|1-2 \sin^2 \frac{\alpha}{2}\right| = 2 \sin \frac{\alpha}{2} + 2 - 4 \sin^2 \frac{\alpha}{2} = \frac{9}{4} - \left(\frac{1}{2} - 2 \sin \frac{\alpha}{2}\right)^2
$$

Since $\sin \frac{\alpha}{2} \in \left[0, \frac{\sqrt{2}}{2}\right]$, we have $E(\alpha) \in [\sqrt{2}, 2]$.

- If $\frac{\alpha}{2} \in \left(\frac{\pi}{4}, \frac{3\pi}{4}\right)$, then

$$
E(\alpha) = 2 \sin \frac{\alpha}{2} + 2\left|1-2 \sin^2 \frac{\alpha}{2}\right| = 2 \sin \frac{\alpha}{2} + 4 \sin^2 \frac{\alpha}{2} - 2 = \left(2 \sin \frac{\alpha}{2} + \frac{1}{2}\right)^2 - \frac{9}{4}
$$

Since $\sin \frac{\alpha}{2} \in \left(\frac{\sqrt{2}}{2}, 1\right)$, we have $E(\alpha) \in (\sqrt{2}, 4)$.

The two cases analyzed say that the expression

$$
E(\alpha, \beta) = \left(2 \sin \frac{\alpha}{2} + 2\left|1-2 \sin^2 \frac{\alpha}{2}\right|\right) + \left(2 \sin \frac{\beta}{2} + 2\left|1-2 \sin^2 \frac{\beta}{2}\right|\right)
$$

has the minimum value $\sqrt{2} + \sqrt{2} = 2\sqrt{2}$, and this is achieved if and only if $\alpha, \beta \in \left\{\frac{\pi}{2}, \frac{3\pi}{2}\right\}$.

Considering (1), we deduce that there are exactly four triples of complex numbers $(u, v)$ that satisfy the hypothesis, namely $(u, v) \in \{(i, i), (-i, i), (i, -i), (-i, -i)\}$.