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Answer:
{∛2e^(iπ), ∛2·e^(iπ/3), ∛2·e^(-iπ/3)}
Step-by-step explanation:
[tex]\displaystyle \sqrt[3]{z}=(2e^{i\pi})^{\frac{1}{3}}=(2e^{i(2n+1)\pi})^{\frac{1}{3}}=\sqrt[3]{2}\cdot e^{i(2n+1)\pi/3}\quad\text{for $n=\{-1,0,1\}$}\\\\=\{\sqrt[3]{2}\cdot e^{-i\pi/3},\sqrt[3]{2}\cdot e^{i\pi/3},\sqrt[3]{2}\cdot e^{i\pi}\}[/tex]
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Additional comment
z = -2. ∛z will be three points in the complex plane on a circle of radius ∛2 at angles ±π/3 and π radians (±60° and 180°). That is, the real cube root of -2 is -∛2.
