## Task B-4.2.

How many complex numbers $z$ are there such that $|z|=1$ and the difference $z^{5!}-z^{4!}$ is a real number?
## Solution.

If $|z|=1$, the number $z$ can be written in trigonometric form as

$$
z=\cos \varphi+i \sin \varphi, \quad 0 \leqslant \varphi<2 \pi
$$

Then,

$$
z^{5!}=\cos (5!\varphi)+i \sin (5!\varphi)=\cos (120 \varphi)+i \sin (120 \varphi)
$$

and

$$
z^{4!}=\cos (4!\varphi)+i \sin (4!\varphi)=\cos (24 \varphi)+i \sin (24 \varphi)
$$

so

$$
z^{5!}-z^{4!}=\cos (120 \varphi)-\cos (24 \varphi)+i(\sin (120 \varphi)-\sin (24 \varphi))
$$

This difference will be a real number if the imaginary part is equal to 0, i.e., if

$$
\sin (120 \varphi)-\sin (24 \varphi)=0
$$

1 point

or

$$
\sin (120 \varphi)=\sin (24 \varphi)
$$

The last equality will be true if the following two possibilities hold:

$$
120 \varphi=24 \varphi+2 k \pi, k \in \mathbb{Z}
$$

or

$$
120 \varphi=\pi-24 \varphi+2 k \pi, k \in \mathbb{Z}
$$

which means

$$
\varphi=\frac{2 k \pi}{96}
$$

or

$$
\varphi=\frac{(2 k+1) \pi}{144}, k \in \mathbb{Z}
$$

Since $0 \leqslant \varphi<2 \pi$ must hold, from the first equality we get

$$
\begin{gathered}
\varphi=\frac{2 k \pi}{96} \Rightarrow 0 \leqslant k<96, k \in \mathbb{Z} \\
\varphi=\frac{(2 k+1) \pi}{144} \Rightarrow-0.5 \leqslant k<143.5, k \in \mathbb{Z}
\end{gathered}
$$

There are $96+144=240$ such integers $k$, which is the number of complex numbers $z$ we are looking for.

Note: Full credit is given if the student solves the equation $\sin (120 \varphi)-\sin (24 \varphi)=0$ by some other method.