Task A-4.3. (8 points)

Calculate $\left(1+\cos \frac{\pi}{7}+i \sin \frac{6 \pi}{7}\right)^{14}$.
## Solution.

First method.

Notice that $1+\cos \frac{\pi}{7}=2 \cos ^{2} \frac{\pi}{14}$ and $\sin \frac{6 \pi}{7}=\sin \frac{\pi}{7}=2 \sin \frac{\pi}{14} \cos \frac{\pi}{14}$. Therefore,

$$
\begin{aligned}
1+\cos \frac{\pi}{7}+i \sin \frac{6 \pi}{7} & =2 \cos ^{2} \frac{\pi}{14}+2 i \sin \frac{\pi}{14} \cos \frac{\pi}{14} \\
& =2 \cos \frac{\pi}{14}\left(\cos \frac{\pi}{14}+i \sin \frac{\pi}{14}\right)
\end{aligned}
$$

By raising to the 14th power, we finally get

$$
\begin{aligned}
\left(1+\cos \frac{\pi}{7}+i \sin \frac{6 \pi}{7}\right)^{14} & =\left(2 \cos \frac{\pi}{14}\left(\cos \frac{\pi}{14}+i \sin \frac{\pi}{14}\right)\right)^{14} \\
& =2^{14}\left(\cos \frac{\pi}{14}\right)^{14}\left(\cos \frac{\pi}{14}+i \sin \frac{\pi}{14}\right)^{14} \\
& =2^{14}\left(\cos \frac{\pi}{14}\right)^{14}(\cos \pi+i \sin \pi) \\
& =-2^{14}\left(\cos \frac{\pi}{14}\right)^{14}
\end{aligned}
$$

Second method.

Transform the number $1+\cos \frac{\pi}{7}+i \sin \frac{6 \pi}{7}$ using sum-to-product formulas:

$$
\begin{aligned}
1+\cos \frac{\pi}{7}+i \sin \frac{6 \pi}{7} & =\cos 0+i \sin 0+\cos \frac{\pi}{7}+i \sin \frac{\pi}{7} \\
& =\left(\cos 0+\cos \frac{\pi}{7}\right)+i\left(\sin 0+\sin \frac{\pi}{7}\right) \\
& =2 \cos \frac{\pi}{14} \cos \frac{\pi}{14}+2 i \sin \frac{\pi}{14} \sin \frac{\pi}{14} \\
& =2 \cos \frac{\pi}{14}\left(\cos \frac{\pi}{14}+i \sin \frac{\pi}{14}\right)
\end{aligned}
$$

We proceed as in the first solution from here.

For 4 points: the given number is expressed in trigonometric form or a good method with a minor calculation error.

For 8 points: the correct result and method.