Solution: Using De Moivre's Theorem, $ z^n = \cos\left(n\theta\right) + i \sin\left(n\theta\right) $. Here, $ z = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) $, so $ z^6 = \cos\left(6 \cdot \frac{\pi}{6}\right) + i \sin\left(6 \cdot \frac{\pi}{6}\right) = \cos(\pi) + i \sin(\pi) $. Evaluating, $ \cos(\pi) = -1 $ and $ \sin(\pi) = 0 $, so $ z^6 = -1 + 0i $.

["Solution Using De Moivre’s Theorem: Evaluating ( z^n ) and Exploring ( z^6 )", "Math students and complex number enthusiasts often encounter elegant ways to compute powers of complex numbers. One powerful tool is De Moivre’s Theorem, which simplifies raising a complex number in polar form to an integer power. In this article, we explore how De Moivre’s Theorem helps evaluate expressions like ( z^n = \cos(n\ heta) + i\sin(n\ heta) ), focusing on a specific example: ( z = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) ). We compute ( z^6 ) step-by-step and uncover its real and imaginary components.", "---", "### Understanding De Moivre’s Theorem", "De Moivre’s Theorem states:", "[\n\left[ \cos(\ heta) + i \sin(\ heta) \right]^n = \cos(n\ heta) + i \sin(n\ heta)\n]", "This powerful identity holds for any real angle ( \ heta ) and integer or rational exponent ( n ). It bridges polar and rectangular forms, making exponentiation of complex numbers intuitive.", "---", "### Applying De Moivre’s Theorem to ( z = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) )", "We begin with:", "[\nz = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right)\n]", "Using De Moivre’s Theorem for ( n = 6 ):", "[\nz^6 = \left[ \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) \right]^6 = \cos\left(6 \cdot \frac{\pi}{6}\right) + i \sin\left(6 \cdot \frac{\pi}{6}\right)\n]", "Simplifying the angle:", "[\n6 \cdot \frac{\pi}{6} = \pi\n]", "Thus:", "[\nz^6 = \cos(\pi) + i \sin(\pi)\n]", "Now evaluate the trigonometric functions:", "- ( \cos(\pi) = -1 )\n- ( \sin(\pi) = 0 )", "So:", "[\nz^6 = -1 + i \cdot 0 = -1\n]", "---", "### Conclusion: ( z^6 = -1 )", "This result confirms that when a complex number lies on the unit circle at angle ( \ heta = \frac{\pi}{6} ), raising it to the 6th power rotates it completely around the origin—6 steps of ( \frac{\pi}{6} ) give ( \pi )—landing at the point (-1) on the real axis.", "De Moivre’s Theorem efficiently handles this rotated power, eliminating the need for direct binomial expansion or manual trigonometric computations.", "---", "### Why This Matters", "Understanding this solution promotes fluency with polar forms and complex arithmetic—key concepts in electrical engineering, signal processing, and quantum physics. Mastering De Moivre’s Theorem unlocks faster, more elegant solutions to power problems involving complex numbers.", "---", "### Summary", "- ( z = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) )\n- By De Moivre’s Theorem:\n [\n z^6 = \cos(\pi) + i \sin(\pi) = -1 + 0i\n ]\n- Final result:\n [\n z^6 = -1\n ]\n- This elegant computation highlights the power of expressing complex numbers in polar form.", "---", "Try it yourself! Use De Moivre’s Theorem to evaluate ( z^{12} ) for the same ( z )—you’ll quickly see ( z^{12} = \cos(2\pi) + i \sin(2\pi) = 1 ).", "---", "Keywords: De Moivre’s Theorem, complex numbers, power of complex numbers, polar form, ( z^n ), ( \cos(n\ heta) + i \sin(n\ heta) ), ( z = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) ), ( z^6 = -1 ), trigonometric identities, mathematics education."]