Question - Finding Third Roots of a Complex Number using De Moivre's Theorem
Solution:
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 $$4\sqrt{3}+4i$$.
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