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Using complex number system Compute the three cube roots of z = −8.​

User Faridghar
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1 Answer

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Write z = -8 in polar form:


z = -8 = 8e^(i\pi)

Then the cube roots of z are


z^(1/3) = 8^(1/3) e^{i\left(\frac{\pi+2n\pi}3\right)

where n ∈ {0, 1, 2}, or


z^(1/3) \in \left\{8^(1/3) e^(i\pi/3), 8^(1/3) e^(i\pi), 8^(1/3) e^(i\,5\pi/3)\right\} \\\\ z^(1/3) \in \left\{2 \left(\frac12 + i\frac{\sqrt3}2\right), -2, 2 \left(\frac12-i\frac{\sqrt3}2\right)\right\} \\\\ \boxed{z^(1/3) \in \left\{1+i\sqrt3, -2, 1-i\sqrt3\right\}}

User Andrew Usikov
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