Cubic Roots Product 1
Let the roots of
\[z^3 = 2 + 2i\]be $ a_1 + ib_1, $ $ a_2 + ib_2, $ and $ a_3 + ib_3 $. Compute $ a_1 a_2 a_3 $.
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- $\frac{a}{b}$
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Solution
Taking the absolute value of both sides, we get
\[|z^3| = |2 + 2i| = 2 \sqrt{2}.\]Then $ |z|^3 = 2 \sqrt{2}, $ so $ |z| = \sqrt{2} $.
Let $ w = \frac{z + \overline{z}}{2}, $ so the possible values of $ w $ are $ a_1, $ $ a_2, $ and $ a_3 $. Then
\[w^3 = \frac{z^3 + 3z^2 \overline{z} + 3z \overline{z}^2 + \overline{z}^3}{8}.\]We know that $ z^3 = 2 + 2i $. Taking the conjugate, we get $ \overline{z^3} = \overline{2 + 2i}, $ so $ \overline{z}^3 = 2 - 2i $. Also,
\[3z^2 \overline{z} + 3z \overline{z} = 3z \overline{z} (z + \overline{z}) = 6|z|^2 w = 12w,\]so
\[w^3 = \frac{2 + 2i + 12w + 2 - 2i}{8} = \frac{4 + 12w}{8} = \frac{3}{2} w + \frac{1}{2}.\]Then
\[w^3 - \frac{3}{2} w - \frac{1}{2} = 0.\]By Vieta's formulas, $ a_1 a_2 a_3 = \boxed{\frac{1}{2}} $.
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