Question

Find all the cube roots of ="z = 2e {}^{i\pi} " alt="z = 2e {}^{i\pi} " align="absmiddle" class="latex-formula">
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Answer 1

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Answer:

  {∛2e^(iπ), ∛2·e^(iπ/3), ∛2·e^(-iπ/3)}

Step-by-step explanation:

  $\displaystyle \sqrt[3]{z}=(2e^{i\pi})^{\frac{1}{3}}=(2e^{i(2n+1)\pi})^{\frac{1}{3}}=\sqrt[3]{2}\cdot e^{i(2n+1)\pi/3}\quad\text{for n=\{-1,0,1\} }\\\\=\{\sqrt[3]{2}\cdot e^{-i\pi/3},\sqrt[3]{2}\cdot e^{i\pi/3},\sqrt[3]{2}\cdot e^{i\pi}\}$

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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.