Questions: Find the cube root of 4-4√3i that graphs in the second quadrant.

Solution Steps

To find the cube root of a complex number that graphs in the second quadrant, we can use De Moivre's Theorem. First, we convert the complex number to its polar form, then apply the cube root, and finally convert it back to rectangular form. We need to ensure the resulting angle places the number in the second quadrant.

Given the complex number \( z = 4 - 4\sqrt{3}i \), we first convert it to polar form. The polar form is given by: \[ z = r (\cos \theta + i \sin \theta) \] where \( r \) is the magnitude and \( \theta \) is the argument (angle).

The magnitude \( r \) is calculated as: \[ r = \sqrt{4^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8 \] The argument \( \theta \) is: \[ \theta = \tan^{-1}\left(\frac{-4\sqrt{3}}{4}\right) = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \]

To find the cube root, we use De Moivre's Theorem: \[ \sqrt[3]{z} = \sqrt[3]{r} \left( \cos \left( \frac{\theta + 2k\pi}{3} \right) + i \sin \left( \frac{\theta + 2k\pi}{3} \right) \right) \] where \( k = 0, 1, 2 \).

The cube root of the magnitude is: \[ \sqrt[3]{r} = \sqrt[3]{8} = 2 \] The argument for the cube root is: \[ \frac{\theta + 2k\pi}{3} = \frac{-\frac{\pi}{3} + 2k\pi}{3} \]

We need to find the root that places the angle in the second quadrant (\(\frac{\pi}{2} < \text{angle} < \pi\)). For \( k = 1 \): \[ \text{angle} = \frac{-\frac{\pi}{3} + 2\pi}{3} = \frac{5\pi}{9} \] This angle is in the second quadrant.

Using the angle \(\frac{5\pi}{9}\) and magnitude 2, we convert back to rectangular form: \[ \sqrt[3]{z} = 2 \left( \cos \frac{5\pi}{9} + i \sin \frac{5\pi}{9} \right) \] Approximating the values: \[ \cos \frac{5\pi}{9} \approx -0.3961, \quad \sin \frac{5\pi}{9} \approx 0.9182 \] Thus: \[ \sqrt[3]{z} \approx 2(-0.3961 + 0.9182i) = -0.7922 + 1.8364i \]

Final Answer