+68 What are the fourth roots of 2√3−2i ? Enter the roots in order of increasing angle measure.
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+506 first convert to polar form
\(2\sqrt{3}-2i\\=4(\frac{\sqrt{3}}{2}-\frac{i}{2})\\=4(\cos(-30)+i\sin(-30))\\=4(cis(-30))\)
The magnitude will be fourth-rooted, while the angle will divide by 4.
Also, note that adding 90 to the angle will still make it a root, because 90*4=360.
Therefore the answers are:
\(\sqrt{2}(cis(-7.5)), \sqrt{2}(cis(-7.5)+90), \sqrt{2}(cis(-7.5+180)), \sqrt{2}(cis(-7.5+270))\)
