Use $z_1 = 4 - 4i$ and $z_2 = \sqrt{3} + i$. Leave answers in polar form. Find $(z_2)^3$.
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$|z_2| = \sqrt{(\sqrt{3})^2 + 1^2} = 2$ $\theta = \arctan(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$ Show more…
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