Identify the Complex Number

The given complex number is $Z = \frac{1}{2} + \frac{\sqrt{3}}{2}i$. We need to find the value of $Z^6$.

Convert Z to Polar Form

First, express the complex number $Z$ in polar form $Z = r(\cos \theta + i \sin \theta)$.

• Modulus (r): $r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$.
• Argument ($\theta$): The argument $\theta$ satisfies $\cos \theta = \frac{1/2}{1} = \frac{1}{2}$ and $\sin \theta = \frac{\sqrt{3}/2}{1} = \frac{\sqrt{3}}{2}$. The principal value for $\theta$ is $\frac{\pi}{3}$ radians.

So, the polar form of $Z$ is $Z = 1 \left(\cos\left(\frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{3}\right)\right)$. This can also be written using Euler's formula as $Z = e^{i\pi/3}$.

Apply De Moivre's Theorem

De Moivre's theorem states that for a complex number $Z = r(\cos \theta + i \sin \theta)$, the $n$-th power is given by $Z^n = r^n (\cos(n\theta) + i \sin(n\theta))$.

In this case, $r=1$, $\theta=\frac{\pi}{3}$, and $n=6$. Applying the theorem:

$Z^6 = 1^6 \left(\cos\left(6 \cdot \frac{\pi}{3}\right) + i \sin\left(6 \cdot \frac{\pi}{3}\right)\right)$

Simplify the Result

Calculate the terms:

$Z^6 = 1 \left(\cos\left(\frac{6\pi}{3}\right) + i \sin\left(\frac{6\pi}{3}\right)\right)$

$Z^6 = \cos(2\pi) + i \sin(2\pi)$

We know that $\cos(2\pi) = 1$ and $\sin(2\pi) = 0$.

$Z^6 = 1 + i \cdot 0$

$Z^6 = 1$

Conclusion

The value of $Z^6$ is 1.