1. Given non-zero real numbers $u, v, w, x, y, z$. How many possible choices of the signs of these numbers are there if

$$
(u+i x)(v+i y)(w+i z)=i ?
$$
Solution. Let's represent the given complex numbers in trigonometric form:

$$
\begin{aligned}
& u+i x=r_{1}\left(\cos \varphi_{1}+i \sin \varphi_{1}\right) \\
& v+i y=r_{2}\left(\cos \varphi_{2}+i \sin \varphi_{2}\right) \\
& w+i z=r_{3}\left(\cos \varphi_{3}+i \sin \varphi_{3}\right)
\end{aligned}
$$

Meanwhile, according to the assumption, $0<\varphi_{i}<2 \pi, \varphi_{i} \neq \frac{k \pi}{2}(k \in \mathbb{Z})$ for $i=1,2,3$. We rewrite equation (1) as:

$$
r_{1} r_{2} r_{3}\left(\cos \left(\varphi_{1}+\varphi_{2}+\varphi_{3}\right)+i \sin \left(\varphi_{1}+\varphi_{2}+\varphi_{3}\right)\right)=\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}
$$

and for a suitable choice of numbers $r_{1}, r_{2}, r_{3}$, it holds if and only if

$$
\varphi_{1}+\varphi_{2}+\varphi_{3}=\frac{\pi}{2}+2 k \pi, k \in\{0,1,2\}
$$

Now we can rephrase the problem as follows:

In a three-dimensional coordinate system, whose axes are labeled $\varphi_{1}, \varphi_{2}, \varphi_{3}$, a cube is given by $0<\varphi_{1}<2 \pi, 0<\varphi_{2}<2 \pi, 0<\varphi_{3}<2 \pi$, which is divided into 64 smaller cubes by the planes $\varphi_{i}=l \frac{\pi}{2}$, ( $i=1,2,3$, $l=1,2,3$ ). We need to determine the number of smaller cubes whose interior has a non-empty intersection with one of the planes $\alpha_{k}$ defined by $\varphi_{1}+\varphi_{2}+\varphi_{3}=\frac{\pi}{2}+2 k \pi, k \in\{0,1,2\}$.
![](https://cdn.mathpix.com/cropped/2024_06_05_18e18734246a9416b6d0g-09.jpg?height=318&width=1240&top_left_y=1593&top_left_x=245)

It is easy to see that the plane $\alpha_{0}$ whose equation is $\varphi_{1}+\varphi_{2}+\varphi_{3}=\frac{\pi}{2}$ intersects the interior of only one cube - the one at the bottom left corner of the large cube (see the diagram). The plane $\alpha_{1}$ whose equation is $\varphi_{1}+\varphi_{2}+\varphi_{3}=\frac{\pi}{2}+2 \pi$ intersects the interior of 7 cubes for which $0<\varphi_{3}<\frac{\pi}{2}$, then 7 cubes for which $\frac{\pi}{2}<\varphi_{3}<\pi$, 5 cubes for which $\pi<\varphi_{3}<\frac{3 \pi}{2}$, and 3 cubes for which $\frac{3 \pi}{2}<\varphi_{3}<2 \pi$, which means a total of 22 cubes. The plane $\alpha_{2}$ whose equation is $\varphi_{1}+\varphi_{2}+\varphi_{3}=\frac{\pi}{2}+4 \pi$ does not intersect the interior of any cube for which $0<\varphi_{3}<\frac{\pi}{2}$, it intersects the interior of one cube for which $\frac{\pi}{2}<\varphi_{3}<\pi$, then 3 cubes for which $\pi<\varphi_{3}<\frac{3 \pi}{2}$, and finally 5 cubes for which $\frac{3 \pi}{2}<\varphi_{3}<2 \pi$, which means a total of 9 cubes.

Thus, the number of cubes sought, and therefore the number of possible choices of signs, is $1+22+9=32$.