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To find the cube roots of [tex]\(-4 + 4 \sqrt{3} i\)[/tex], we can use De Moivre's theorem for complex numbers, which is ideal for this purpose. Here's the step-by-step procedure:
1. Convert the complex number to polar form:
- A complex number [tex]\( z = a + bi \)[/tex] can be written in polar form as [tex]\( z = r(\cos \theta + i \sin \theta) \)[/tex], where [tex]\( r \)[/tex] is the magnitude (modulus) and [tex]\( \theta \)[/tex] is the argument (angle).
- For [tex]\( z = -4 + 4 \sqrt{3} i \)[/tex]:
[tex]\[ r = |z| = \sqrt{(-4)^2 + (4 \sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8 \][/tex]
[tex]\[ \theta = \tan^{-1}\left(\frac{\text{Imaginary part}}{\text{Real part}}\right) = \tan^{-1}\left(\frac{4 \sqrt{3}}{-4}\right) = \tan^{-1}(-\sqrt{3}) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \][/tex]
2. Divide the argument by 3 to find the angles for the cube roots:
- The cube roots will be at angles [tex]\( \frac{\theta + 2k\pi}{3} \)[/tex] for [tex]\( k = 0, 1, 2 \)[/tex].
3. Calculate each of the angles:
- For [tex]\( k = 0 \)[/tex]:
[tex]\[ \theta_0 = \frac{\frac{2\pi}{3} + 2 \cdot 0 \cdot \pi}{3} = \frac{2\pi}{9} \][/tex]
- For [tex]\( k = 1 \)[/tex]:
[tex]\[ \theta_1 = \frac{\frac{2\pi}{3} + 2 \cdot \pi}{3} = \frac{\frac{2\pi}{3} + 2\pi}{3} = \frac{8\pi}{9} \][/tex]
- For [tex]\( k = 2 \)[/tex]:
[tex]\[ \theta_2 = \frac{\frac{2\pi}{3} + 4\pi}{3} = \frac{\frac{2\pi}{3} + 6\pi}{3} = \frac{14\pi}{9} \][/tex]
4. Compute the cube roots using the magnitude [tex]\( r^{1/3} \)[/tex] and the angles:
- The magnitude for each root is [tex]\( 8^{1/3} = 2 \)[/tex].
Thus, the cube roots are:
[tex]\[ z_1 = 2 \left(\cos \frac{2\pi}{9} + i \sin \frac{2\pi}{9}\right) \][/tex]
[tex]\[ z_2 = 2 \left(\cos \frac{8\pi}{9} + i \sin \frac{8\pi}{9}\right) \][/tex]
[tex]\[ z_3 = 2 \left(\cos \frac{14\pi}{9} + i \sin \frac{14\pi}{9}\right) \][/tex]
Therefore, the cube roots of [tex]\(-4 + 4 \sqrt{3} i\)[/tex] are:
[tex]\[ 2\left(\cos \frac{2 \pi}{9}+i \sin \frac{2 \pi}{9}\right) \][/tex]
[tex]\[ 2\left(\cos \frac{8 \pi}{9}+i \sin \frac{8 \pi}{9}\right) \][/tex]
[tex]\[ 2\left(\cos \frac{14 \pi}{9}+i \sin \frac{14 \pi}{9}\right) \][/tex]
Among the given options, the correct cube roots are:
[tex]\[ 2\left(\cos \frac{8 \pi}{9}+i \sin \frac{8 \pi}{9}\right) \][/tex]
[tex]\[ 2\left(\cos \frac{14 \pi}{9}+i \sin \frac{14 \pi}{9}\right) \][/tex]
[tex]\[ 2\left(\cos \frac{2 \pi}{9}+i \sin \frac{2 \pi}{9}\right) \][/tex]
Thus, the correct answers are:
[tex]\[ \boxed{2\left(\cos \frac{2 \pi}{9}+i \sin \frac{2 \pi}{9}\right)}, \boxed{2\left(\cos \frac{8 \pi}{9}+i \sin \frac{8 \pi}{9}\right)}, \boxed{2\left(\cos \frac{14 \pi}{9}+i \sin \frac{14 \pi}{9}\right)} \][/tex]
1. Convert the complex number to polar form:
- A complex number [tex]\( z = a + bi \)[/tex] can be written in polar form as [tex]\( z = r(\cos \theta + i \sin \theta) \)[/tex], where [tex]\( r \)[/tex] is the magnitude (modulus) and [tex]\( \theta \)[/tex] is the argument (angle).
- For [tex]\( z = -4 + 4 \sqrt{3} i \)[/tex]:
[tex]\[ r = |z| = \sqrt{(-4)^2 + (4 \sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8 \][/tex]
[tex]\[ \theta = \tan^{-1}\left(\frac{\text{Imaginary part}}{\text{Real part}}\right) = \tan^{-1}\left(\frac{4 \sqrt{3}}{-4}\right) = \tan^{-1}(-\sqrt{3}) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \][/tex]
2. Divide the argument by 3 to find the angles for the cube roots:
- The cube roots will be at angles [tex]\( \frac{\theta + 2k\pi}{3} \)[/tex] for [tex]\( k = 0, 1, 2 \)[/tex].
3. Calculate each of the angles:
- For [tex]\( k = 0 \)[/tex]:
[tex]\[ \theta_0 = \frac{\frac{2\pi}{3} + 2 \cdot 0 \cdot \pi}{3} = \frac{2\pi}{9} \][/tex]
- For [tex]\( k = 1 \)[/tex]:
[tex]\[ \theta_1 = \frac{\frac{2\pi}{3} + 2 \cdot \pi}{3} = \frac{\frac{2\pi}{3} + 2\pi}{3} = \frac{8\pi}{9} \][/tex]
- For [tex]\( k = 2 \)[/tex]:
[tex]\[ \theta_2 = \frac{\frac{2\pi}{3} + 4\pi}{3} = \frac{\frac{2\pi}{3} + 6\pi}{3} = \frac{14\pi}{9} \][/tex]
4. Compute the cube roots using the magnitude [tex]\( r^{1/3} \)[/tex] and the angles:
- The magnitude for each root is [tex]\( 8^{1/3} = 2 \)[/tex].
Thus, the cube roots are:
[tex]\[ z_1 = 2 \left(\cos \frac{2\pi}{9} + i \sin \frac{2\pi}{9}\right) \][/tex]
[tex]\[ z_2 = 2 \left(\cos \frac{8\pi}{9} + i \sin \frac{8\pi}{9}\right) \][/tex]
[tex]\[ z_3 = 2 \left(\cos \frac{14\pi}{9} + i \sin \frac{14\pi}{9}\right) \][/tex]
Therefore, the cube roots of [tex]\(-4 + 4 \sqrt{3} i\)[/tex] are:
[tex]\[ 2\left(\cos \frac{2 \pi}{9}+i \sin \frac{2 \pi}{9}\right) \][/tex]
[tex]\[ 2\left(\cos \frac{8 \pi}{9}+i \sin \frac{8 \pi}{9}\right) \][/tex]
[tex]\[ 2\left(\cos \frac{14 \pi}{9}+i \sin \frac{14 \pi}{9}\right) \][/tex]
Among the given options, the correct cube roots are:
[tex]\[ 2\left(\cos \frac{8 \pi}{9}+i \sin \frac{8 \pi}{9}\right) \][/tex]
[tex]\[ 2\left(\cos \frac{14 \pi}{9}+i \sin \frac{14 \pi}{9}\right) \][/tex]
[tex]\[ 2\left(\cos \frac{2 \pi}{9}+i \sin \frac{2 \pi}{9}\right) \][/tex]
Thus, the correct answers are:
[tex]\[ \boxed{2\left(\cos \frac{2 \pi}{9}+i \sin \frac{2 \pi}{9}\right)}, \boxed{2\left(\cos \frac{8 \pi}{9}+i \sin \frac{8 \pi}{9}\right)}, \boxed{2\left(\cos \frac{14 \pi}{9}+i \sin \frac{14 \pi}{9}\right)} \][/tex]
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