We are tasked with finding $(\overline{z})^4$, where $z = \frac{1}{3} + \frac{\sqrt{3}}{2} i$, and $\overline{z}$ represents the complex conjugate of $z$. Let's break this down step by step:

Step 1: Identify the complex conjugate of $z$

Given $z = \frac{1}{3} + \frac{\sqrt{3}}{2} i$, its complex conjugate $\overline{z}$ is found by changing the sign of the imaginary part: $\overline{z} = \frac{1}{3} - \frac{\sqrt{3}}{2} i$

Step 2: Express $\overline{z}$ in polar form

To simplify powers of complex numbers, it's easier to convert $\overline{z}$ into polar form. We need to calculate the modulus $r$ and argument $\theta$.

Modulus of $\overline{z}$:

$r = \left| \overline{z} \right| = \sqrt{\left( \frac{1}{3} \right)^2 + \left( -\frac{\sqrt{3}}{2} \right)^2}$ $r = \sqrt{\frac{1}{9} + \frac{3}{4}} = \sqrt{\frac{1}{9} + \frac{27}{36}} = \sqrt{\frac{4}{12}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$

Argument $\theta$:

We find $\theta$ by considering the arctangent of the imaginary and real parts: $\theta = \tan^{-1} \left( \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{3}} \right) = \tan^{-1} \left( -\frac{\sqrt{3}}{2} \times 3 \right) = \tan^{-1} (-\frac{3\sqrt{3}}{2}) \approx -\frac{\pi}{3}$ Thus, $\overline{z} = r (\cos \theta + i \sin \theta)$, where $r = \frac{2\sqrt{3}}{3}$ and $\theta = -\frac{\pi}{3}$.

Step 3: Raise $\overline{z}$ to the fourth power

In polar form, raising a complex number to a power simplifies to raising the modulus to the power and multiplying the argument by the power: $(\overline{z})^4 = r^4 \left( \cos(4\theta) + i \sin(4\theta) \right)$

• $r^4 = \left( \frac{2\sqrt{3}}{3} \right)^4 = \frac{16 \cdot 3^2}{81} = \frac{144}{81} = \frac{16}{9}$
• $4\theta = 4 \times -\frac{\pi}{3} = -\frac{4\pi}{3}$

Now, use the periodicity of cosine and sine to simplify $\cos \left( -\frac{4\pi}{3} \right)$ and $\sin \left( -\frac{4\pi}{3} \right)$: $\cos \left( -\frac{4\pi}{3} \right) = -\frac{1}{2}, \quad \sin \left( -\frac{4\pi}{3} \right) = \frac{\sqrt{3}}{2}$

Step 4: Final result

$(\overline{z})^4 = \frac{16}{9} \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right)$ $(\overline{z})^4 = \frac{16}{9} \times \left( -\frac{1}{2} \right) + i \frac{16}{9} \times \frac{\sqrt{3}}{2}$ $(\overline{z})^4 = -\frac{8}{9} + i \frac{8\sqrt{3}}{9}$

Thus, $(\overline{z})^4 = -\frac{8}{9} + i \frac{8\sqrt{3}}{9}$.

Would you like more details or have any questions about this? Here are five related questions:

1. How do you convert any complex number into polar form?
2. What is the geometric interpretation of complex conjugates?
3. How does De Moivre's Theorem help in raising complex numbers to powers?
4. Can you explain why the modulus of a complex number remains the same for both $z$ and $\overline{z}$?
5. How can we verify the result of powers of complex numbers using direct multiplication?

Tip: When dealing with powers of complex numbers, polar form greatly simplifies the calculations, especially when using De Moivre's Theorem.