Question

Find all the cube roots of

`z = 2e {}^(i\pi)`
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Answer 1

9514 1404 393

Answer:

{∛2e^(iπ), ∛2·e^(iπ/3), ∛2·e^(-iπ/3)}

Explanation:

`\displaystyle \sqrt[3]{z}=(2e^(i\pi))^{(1)/(3)}=(2e^(i(2n+1)\pi))^{(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)\}`

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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.