Finding Third Roots of a Complex Number using De Moivre's Theorem
The image displays a math problem that asks to find the third roots of the complex number \(4\sqrt{3}+4i\). The third roots of a complex number can be found using De Moivre's Theorem, which states that for a complex number in polar form \(r(\cos \theta + i\sin \theta)\), its n-th roots are given by:
\[ r^{1/n} \left( \cos \frac{\theta + 2k\pi}{n} + i\sin \frac{\theta + 2k\pi}{n} \right) \]
where \(k = 0, 1, 2, \ldots, n-1\).
To apply De Moivre's Theorem, we first need to express the given complex number \(4\sqrt{3}+4i\) in polar form, which is \(r(\cos \theta + i\sin \theta)\), where \(r\) is the magnitude of the complex number and \(\theta\) is the argument (angle).
1. Compute the magnitude \(r\):
\[ r = \sqrt{(4\sqrt{3})^2 + (4)^2} = \sqrt{48 + 16} = \sqrt{64} = 8 \]
2. Determine the argument \(\theta\):
\[ \cos \theta = \frac{\text{Real part}}{r} = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2} \]
\[ \sin \theta = \frac{\text{Imaginary part}}{r} = \frac{4}{8} = \frac{1}{2} \]
Since the complex number is in the first quadrant, \(\theta\) is \(30^\circ\) or \(\frac{\pi}{6}\) radians.
3. Use De Moivre's Theorem to find the third roots:
For \(k = 0\):
\[ \sqrt[3]{8} \left( \cos \frac{\pi/6 + 2(0)\pi}{3} + i\sin \frac{\pi/6 + 2(0)\pi}{3} \right) = 2 \left( \cos \frac{\pi}{18} + i\sin \frac{\pi}{18} \right) \]
For \(k = 1\):
\[ \sqrt[3]{8} \left( \cos \frac{\pi/6 + 2(1)\pi}{3} + i\sin \frac{\pi/6 + 2(1)\pi}{3} \right) = 2 \left( \cos \frac{7\pi}{18} + i\sin \frac{7\pi}{18} \right) \]
For \(k = 2\):
\[ \sqrt[3]{8} \left( \cos \frac{\pi/6 + 2(2)\pi}{3} + i\sin \frac{\pi/6 + 2(2)\pi}{3} \right) = 2 \left( \cos \frac{13\pi}{18} + i\sin \frac{13\pi}{18} \right) \]
These are the three third roots of the complex number \(4\sqrt{3}+4i\).
