266. Calculate $z=\sqrt[4]{-16}$.
Solution. Let's write the number -16 in trigonometric form. We find \( r = \sqrt{a^2 + b^2} = \sqrt{(-16)^2 + 0} = 16 \). Here, \( a = -16 \), \( b = 0 \), and thus the point is located on the negative part of the \( O x \) axis; therefore, \( \varphi = \pi \).

To find the roots, we use the formula

\[
\sqrt[4]{z} = \sqrt[4]{16} \left( \cos \frac{\pi + 2\pi k}{4} + i \sin \frac{\pi + 2\pi k}{4} \right)
\]

where \( k = 0, k = 1, k = 2 \), and \( k = 3 \). Accordingly, we get

\[
z^{(1)} = \sqrt[4]{16} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) = 2 \left( \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = \sqrt{2} + i \sqrt{2}
\]

\[
\begin{gathered}
z^{(2)} = \sqrt[4]{16} \left( \cos \frac{\pi + 2\pi}{4} + i \sin \frac{2\pi + \pi}{4} \right) = 2 \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right) = \\
= -\sqrt{2} + i \sqrt{2} \\
= \\
\begin{array}{c}
z^{(3)} = \sqrt[4]{16} \left( \cos \frac{\pi + 4\pi}{4} + i \sin \frac{\pi + 4\pi}{4} \right) = 2 \left( \cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4} \right) = \\
z^{(4)} = \sqrt{2} - i \sqrt{2}
\end{array} \\
= \sqrt{16} \left( \cos \frac{\pi + 6\pi}{4} + i \sin \frac{\pi + 6\pi}{4} \right) = 2 \left( \cos \frac{7\pi}{4} + i \sin \frac{7\pi}{4} \right) =
\end{gathered}
\]

267-273. Perform the operations, first writing the complex numbers in trigonometric form: