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find all the cube roots of -8+8i(sqrt)3

Sagot :

Step-by-step explanation:

We need to find cube roots of

[tex] - 8 + 8i \sqrt{3} [/tex]

Let put this into trig form.

[tex]r( \cos(x) + i \: sin(x)[/tex]

To find the radius we do this formula,

[tex]r = \sqrt{ {a}^{2} + {b}^{2} } [/tex]

where a is the real number and b is the coeffeicent of the imaginary term.

[tex]r = \sqrt{ - 8 {}^{2} + (8 \sqrt{3} ) {}^{2} } = \sqrt{64 + 192} = \sqrt{256} = 16[/tex]

So our r value 16.

We need to find x, so we do this rule.

[tex] \tan(x) = \frac{b}{a} [/tex]

[tex] \tan(x) = - \sqrt{3} [/tex]

[tex](x) = \frac{2\pi}{3} [/tex]

So our x is

[tex]x = \frac{2\pi}{3} [/tex]

Now, we can simply our trig form,

[tex]16 ( \cos( \frac{2\pi}{3} ) + \sin( \frac{2\pi}{3} ) )[/tex]

Now, we must find the cube roots of this.

This what we do,

Step 1: Take the cube root of 16., which is

[tex]2 \sqrt[3]{2} [/tex]

Step 2: Divide the x values, which 2pi/3 by 3.

So now we get

[tex] \frac{ \frac{2\pi}{3} }{3} = \frac{2\pi}{9} [/tex]

Step 3: Divide 2 pi by 3 and add that 2 times to 2pi/9.

[tex] \frac{2\pi}{9} + \frac{2\pi}{3} + \frac{2\pi}{3} = \frac{14\pi}{9} [/tex]

So our cube root of this is

[tex] 2 \sqrt[3]{2} ( \cos \frac{14\pi}{9} + i \: sin \: \frac{14\pi}{9} ) [/tex]

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