Answer:
The answer is
[tex]2 \cos(100) + i \sin(100) [/tex]
Step-by-step explanation:
This is a complex number,
[tex]a + bi[/tex]
First, convert this to de movire form.
[tex]r( \cos( \alpha ) + i \sin( \alpha ) [/tex]
where
[tex]r = \sqrt{ {a}^{2} + {b}^{2} } [/tex]
and
[tex] \alpha = \tan {}^{ - 1} ( \frac{b}{a} ) [/tex]
[tex]a = 4[/tex]
[tex]b = - 4 \sqrt{ 3} i[/tex]
[tex]r = \sqrt{ {4}^{2} + ( - 4 \sqrt{3}) {}^{2} } [/tex]
[tex]r = \sqrt{16 + 48} [/tex]
[tex]r = \sqrt{64} = 8[/tex]
and
[tex] \alpha = \tan {}^{ - 1} ( \frac{ - 4 \sqrt{3} }{4} ) [/tex]
[tex] \alpha = \tan {}^{ - 1} ( - \sqrt{3} ) [/tex]
Here, our a is positive and b is negative so our angle in degrees must lie in the fourth quadrant, that angle is 300 degrees.
So
[tex] \alpha = 300[/tex]
So our initially form is
[tex]8( \cos(300) + i \sin(300) )[/tex]
Now, we use the roots of unity formula. To do this, we first take the cube root of the modulus, 8,
[tex] \sqrt[3]{8} = 2[/tex]
Next, since cos and sin have a period of 360 we add 360 to each degree then we divide it by 3.
[tex] \sqrt[3]{8} ( \cos( \frac{300 + 360n}{3} ) + \sin( \frac{300 + 360n}{3} ) [/tex]
[tex]2 \cos(100 + 120n) + i \sin(100 + 120n) [/tex]
Since 100 is in the second quadrant, we let n=0,
[tex]2 \cos(100) + i \sin(100) [/tex]
