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

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Answer from Sia

Posted 6 months ago

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) $\theta$ of the complex number is $\arctan\left(\frac{8\sqrt{3}}{-8}\right) = \arctan(-\sqrt{3})$, 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 $16(\cos(\theta) + i\sin(\theta))$. Using the two possible values for $\theta$, 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 $4\left(-\frac{\sqrt{3}}{2} + \frac{1}{2}i\right)$ and $4\left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)$

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.

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