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The expression you've uploaded appears to be:

\[
\left(-1 - \sqrt{3}i\right)^{\frac{1}{4}}
\]

This represents the fourth root of a complex number. To solve this, we need to convert the complex number into polar form and apply De Moivre's Theorem. Here's how we can break it down:

### 1. Convert to Polar Form
A complex number \( z = a + bi \) can be represented in polar form as \( r (\cos \theta + i \sin \theta) \), where:

- \( r = \sqrt{a^2 + b^2} \) is the modulus.
- \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \) is the argument.

For \( z = -1 - \sqrt{3}i \):
- \( a = -1 \)
- \( b = -\sqrt{3} \)

#### Modulus \( r \):
\[
r = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2
\]

#### Argument \( \theta \):
\[
\theta = \tan^{-1}\left(\frac{-\sqrt{3}}{-1}\right) = \tan^{-1}(\sqrt{3})
\]
Since the point is in the third quadrant (both real and imaginary parts are negative), the argument is \( \pi + \frac{\pi}{3} = \frac{4\pi}{3} \).

Thus, in polar form:
\[
z = 2 \left(\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}\right)
\]

### 2. Apply De Moivre's Theorem
To take the fourth root, we apply De Moivre's Theorem. The fourth roots of a complex number in polar form are given by:

\[
z_k = r^{1/n} \left(\cos \frac{\theta + 2k\pi}{n} + i \sin \frac{\theta + 2k\pi}{n}\right)
\]

Where \( n = 4 \) (since we're taking the fourth root) and \( k = 0, 1, 2, 3 \) (because we get four distinct roots).

#### Fourth root of the modulus:
\[
r^{1/4} = 2^{1/4} = \sqrt[4]{2}
\]

#### Roots:
The argument for each root will be:
\[
\frac{\frac{4\pi}{3} + 2k\pi}{4} = \frac{\pi}{3} + \frac{k\pi}{2}
\]
So the four roots will have arguments:
- For \( k = 0 \): \( \frac{\pi}{3} \)
- For \( k = 1 \): \( \frac{\pi}{3} + \frac{\pi}{2} = \frac{5\pi}{6} \)
- For \( k = 2 \): \( \frac{\pi}{3} + \pi = \frac{4\pi}{3} \)
- For \( k = 3 \): \( \frac{\pi}{3} + \frac{3\pi}{2} = \frac{11\pi}{6} \)

Thus, the four roots are:
\[
\sqrt[4]{2} \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)
\]
\[
\sqrt[4]{2} \left( \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \right)
\]
\[
\sqrt[4]{2} \left( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right)
\]
\[
\sqrt[4]{2} \left( \cos \frac{11\pi}{6} + i \sin \frac{11\pi}{6} \right)
\]

These are the four fourth roots of the complex number \( -1 - \sqrt{3}i \).

Would you like further details on any of these steps or a more specific part of the solution?

### Relative Questions:
1. How do you convert a complex number from rectangular to polar form?
2. What is the geometric significance of taking roots of a complex number?
3. How is De Moivre’s Theorem used in finding complex roots?
4. Can you explain why there are four distinct roots for a fourth root of a complex number?
5. How do the arguments of a complex number's roots differ?

### Tip:
When dealing with complex numbers in polar form, always remember to consider the quadrant to correctly determine the argument angle \( \theta \).

Explore how to find the fourth root of the complex number (-1 - √3i) using De Moivre's Theorem. This problem involves converting the number into polar form, calculating the modulus and argument, and applying De Moivre's Theorem to determine the roots. Suitable for advanced mathematics students in Grades 11-12.

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