Calculate $\sqrt[3]{- i}$

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Solution

Let $z = - i = e^{- i π / 2}$
$z^{1 / 3} = (- i)^{1 / 3} = e^{- i π / 6}$
$⟹ (cos - π / 6 + i sin - π / 6)$
$⟹ \frac{\sqrt{3}}{2} - \frac{i}{2}$

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Similar questions

i = \sqrt{- 1}

(3 + i) (3 - i)

u = 1 + i \sqrt{3}

v = \sqrt{3} + i

Calculate $\frac{u^{3}}{v^{4}}$

i^{2} = - 1

3 i^{2} + i^{3} - i^{4}

\sqrt[4]{- 1 \frac{1}{2} - i \frac{\sqrt{3}}{2}} .

I_{1}, I_{2}, I_{3}, I_{4}

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