## Task B-4.6.

Let $z=(1-i \sqrt{3})^{p}$ and $w=(1+i)^{q}$. Determine the smallest natural numbers $p$ and $q$ such that the complex numbers $z$ and $w$ are equal.
## First Solution.

Let's write the complex numbers $1-i \sqrt{3}, 1+i$ in trigonometric form:

$$
\begin{aligned}
1-i \sqrt{3} & =2\left(\cos \frac{5 \pi}{3}+i \sin \frac{5 \pi}{3}\right) \\
1+i & =\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)
\end{aligned}
$$

Then we have

$$
\begin{aligned}
& z=2^{p}\left(\cos \frac{5 p \pi}{3}+i \sin \frac{5 p \pi}{3}\right) \\
& w=\sqrt{2}^{q}\left(\cos \frac{q \pi}{4}+i \sin \frac{q \pi}{4}\right)
\end{aligned}
$$

If $z=w$, then $2^{p}=\sqrt{2}^{q}$ and $\frac{5 p \pi}{3}=\frac{q \pi}{4}+2 k \pi, k \in \mathbb{Z}$.

From $2^{p}=\sqrt{2}^{q}$, or $2^{p}=2^{\frac{q}{2}}$, it follows that $q=2 p$.

$$
\begin{gathered}
\frac{5 p \pi}{3}-\frac{q \pi}{4}=2 k \pi, k \in \mathbb{Z} \\
\frac{5 p \pi}{3}-\frac{2 p \pi}{4}=2 k \pi, k \in \mathbb{Z} \\
\frac{7 p \pi}{6}=2 k \pi, k \in \mathbb{Z} \\
p=\frac{12}{7} k, k \in \mathbb{Z}
\end{gathered}
$$

Thus, the smallest value of $k$ for which $p$ is a natural number is 7.

Then $p=12, q=24$.

## Second Solution.

Notice that from $|z|=|w|$ it follows that $2^{p}=(\sqrt{2})^{q}$. Then $2 p=q$.

We are looking for $p$ such that $(1-i \sqrt{3})^{p}=(1+i)^{2 p}$, or,

$(1-i \sqrt{3})^{p}=(2 i)^{p},\left(\frac{1-i \sqrt{3}}{2}\right)^{p}=i^{p}$.

The left side can have only 6 values:

$1, \frac{1}{2}-i \frac{\sqrt{3}}{2},-\frac{1}{2}-i \frac{\sqrt{3}}{2},-1,-\frac{1}{2}+i \frac{\sqrt{3}}{2}, \frac{1}{2}+i \frac{\sqrt{3}}{2}$,

(for $p \in\{6 m, 6 m+1,6 m+2,6 m+3,6 m+4,6 m+5\}, m \in \mathbb{N}$ ).

The right side can have 4 values $1, i,-1,-i$ (for $p \in\{4 l, 4 l+1,4 l+2,4 l+3\}, l \in$ $\mathbb{N}$ ).

Since the left and right sides must be equal, $p$ is such that both are equal to 1 or both are equal to -1.

In the first case, $p=6 m=4 l$, so $p$ is divisible by 3 and 4, hence $p=12 k, k \in \mathbb{N}$.

In the second case, $p=6 m+3=4 l+2$ which is impossible because $p$ would be both even and odd at the same time.

Finally, the solution consists of all numbers $p$ divisible by 12. The smallest sought numbers are $p=12$, $q=24$.