6) Find the complex square roots of $-8+8 i \sqrt{3}$. Give answers in rectangular form.

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∻Solution by Steps∻ ‖ step 1 ⋮ To find the complex square roots of $-8+8i\sqrt{3}$, we first express the complex number in polar form. ‖ ‖ step 2 ⋮ The magnitude of the complex number is $\sqrt{(-8)^2+(8\sqrt{3})^2} = \sqrt{64+192} = \sqrt{256} = 16$. ‖ ‖ step 3 ⋮ The argument (angle) $ heta$ of the complex number is $\arctan\left(\frac{8\sqrt{3}}{-8} ight) = \arctan(-\sqrt{3})$, which is $\frac{5\pi}{3}$ or $\frac{2\pi}{3}$, considering the complex number is in the second quadrant. ‖ ‖ step 4 ⋮ The polar form is $16(\cos( heta) + i\sin( heta))$. Using the two possible values for $ heta$, we have $16(\cos(\frac{5\pi}{3}) + i\sin(\frac{5\pi}{3}))$ and $16(\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3}))$. ‖ ‖ step 5 ⋮ To find the square roots, we take the square root of the magnitude and halve the angles. The magnitude of the square roots is $\sqrt{16} = 4$. ‖ ‖ step 6 ⋮ The angles for the square roots are $\frac{5\pi}{6}$ and $\frac{\pi}{3}$. ‖ ‖ step 7 ⋮ The two complex square roots in polar form are $4(\cos(\frac{5\pi}{6}) + i\sin(\frac{5\pi}{6}))$ and $4(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))$. ‖ ‖ step 8 ⋮ Converting these back to rectangular form, we get $4\left(-\frac{\sqrt{3}}{2} + \frac{1}{2}i ight)$ and $4\left(\frac{1}{2} + \frac{\sqrt{3}}{2}i ight)$. ‖ ‖ step 9 ⋮ Simplifying, we obtain the complex square roots: $-2\sqrt{3} + 2i$ and $2 + 2\sqrt{3}i$. ‖ ∻Answer∻ ⚹ $-2\sqrt{3} + 2i$ and $2 + 2\sqrt{3}i$ ⚹ ∻Key Concept∻ ⚹Finding complex square roots⚹ ∻Explanation∻ ⚹To find the square roots of a complex number, we convert it to polar form, take the square root of the magnitude, and halve the angle. Then we convert the result back to rectangular form.⚹◊What is the formula to find the complex square roots of a number in rectangular form?⍭ Generate me a similar question◊

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