We are tasked with solving the equation \( Z^4 = \frac{1 - i}{1 + i\sqrt{3}} \).

### Step 1: Express the right-hand side in polar form
First, we want to express the complex fraction on the right-hand side in polar form. To do this, let's break down the numerator and denominator:

- **Numerator**: \( 1 - i \)
  - The modulus of \( 1 - i \) is given by:
    \[
    |1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}
    \]
  - The argument of \( 1 - i \) (in radians) is:
    \[
    \arg(1 - i) = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4}
    \]
  Therefore, in polar form, \( 1 - i = \sqrt{2} e^{-i\pi/4} \).

- **Denominator**: \( 1 + i\sqrt{3} \)
  - The modulus of \( 1 + i\sqrt{3} \) is:
    \[
    |1 + i\sqrt{3}| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2
    \]
  - The argument of \( 1 + i\sqrt{3} \) is:
    \[
    \arg(1 + i\sqrt{3}) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}
    \]
  Therefore, in polar form, \( 1 + i\sqrt{3} = 2 e^{i\pi/3} \).

### Step 2: Simplify the complex fraction
Now we rewrite the right-hand side of the equation using these polar forms:
\[
\frac{1 - i}{1 + i\sqrt{3}} = \frac{\sqrt{2} e^{-i\pi/4}}{2 e^{i\pi/3}} = \frac{\sqrt{2}}{2} e^{i\left(-\frac{\pi}{4} - \frac{\pi}{3}\right)}
\]
Simplify the exponent:
\[
-\frac{\pi}{4} - \frac{\pi}{3} = -\frac{3\pi}{12} - \frac{4\pi}{12} = -\frac{7\pi}{12}
\]
Thus, the equation becomes:
\[
Z^4 = \frac{\sqrt{2}}{2} e^{-i7\pi/12}
\]

### Step 3: Solve for \( Z \)
Now, we take the fourth root of both sides. To find the fourth roots of a complex number in polar form, we use the formula:
\[
Z_k = r^{1/4} e^{i \left(\frac{\theta + 2k\pi}{4}\right)}
\]
where \( r = \frac{\sqrt{2}}{2} \) and \( \theta = -\frac{7\pi}{12} \), and \( k = 0, 1, 2, 3 \).

- The modulus of \( Z_k \) is:
  \[
  r^{1/4} = \left(\frac{\sqrt{2}}{2}\right)^{1/4} = 2^{-1/8}
  \]

- The argument of \( Z_k \) is:
  \[
  \frac{-\frac{7\pi}{12} + 2k\pi}{4} = \frac{-7\pi + 24k\pi}{48} = \frac{(24k - 7)\pi}{48}
  \]

Thus, the four solutions are:
\[
Z_k = 2^{-1/8} e^{i\frac{(24k - 7)\pi}{48}}, \quad k = 0, 1, 2, 3
\]

### Final Answer:
The four solutions for \( Z \) are:
\[
Z_0 = 2^{-1/8} e^{i\frac{-7\pi}{48}}, \quad Z_1 = 2^{-1/8} e^{i\frac{17\pi}{48}}, \quad Z_2 = 2^{-1/8} e^{i\frac{41\pi}{48}}, \quad Z_3 = 2^{-1/8} e^{i\frac{65\pi}{48}}
\]

Would you like me to expand on any of the steps or explain further? 

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**Related Questions:**
1. How do you convert a complex number into polar form?
2. What is the significance of taking roots of complex numbers in polar form?
3. How do you find the argument of a complex number geometrically?
4. What are the practical applications of finding the roots of complex numbers?
5. How do you perform multiplication and division of complex numbers in polar form?

**Tip**: Always convert complex numbers to polar form for easier multiplication, division, and root extraction!

Learn how to solve the complex equation Z⁴ = (1-i)/(1+i√3) by converting the right-hand side to polar form and using De Moivre's Theorem to find the four solutions. This solution involves key concepts in complex numbers and polar form, suitable for advanced high school students.

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