Let's solve \( (-1 - i)^9 + (-4 + 2i)^6 \) step-by-step.

### Step 1: Compute \( (-1 - i)^9 \)

1. Convert \( -1 - i \) to polar form. First, find the modulus and argument:
   - The modulus \( r \) is \( | -1 - i | = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2} \).
   - The argument \( \theta \) (in radians) is \( \arctan\left(\frac{-1}{-1}\right) - \pi = -\frac{3\pi}{4} \) since the point is in the third quadrant.

2. In polar form, \( -1 - i = \sqrt{2} \cdot e^{i \cdot (-\frac{3\pi}{4})} \).

3. Raise this to the power 9:
   \[
   (-1 - i)^9 = \left(\sqrt{2} \cdot e^{i \cdot (-\frac{3\pi}{4})}\right)^9 = (\sqrt{2})^9 \cdot e^{i \cdot (-\frac{27\pi}{4})}
   \]
   - Simplify the modulus: \( (\sqrt{2})^9 = 2^{4.5} = 16\sqrt{2} \).
   - For the argument, reduce \( -\frac{27\pi}{4} \) to an equivalent angle within \( [0, 2\pi) \): \( -\frac{27\pi}{4} + 7\pi = \frac{5\pi}{4} \).

   Therefore:
   \[
   (-1 - i)^9 = 16\sqrt{2} \cdot e^{i \cdot \frac{5\pi}{4}}
   \]
   In rectangular form:
   \[
   = -16 - 16i
   \]

### Step 2: Compute \( (-4 + 2i)^6 \)

1. Convert \( -4 + 2i \) to polar form. First, find the modulus and argument:
   - The modulus \( r \) is \( | -4 + 2i | = \sqrt{(-4)^2 + 2^2} = \sqrt{20} = 2\sqrt{5} \).
   - The argument \( \theta \) is \( \arctan\left(\frac{2}{-4}\right) + \pi = \frac{\pi}{2} \) (second quadrant).

2. In polar form, \( -4 + 2i = 2\sqrt{5} \cdot e^{i \cdot \frac{\pi}{2}} \).

3. Raise this to the power 6:
   \[
   (-4 + 2i)^6 = \left(2\sqrt{5} \cdot e^{i \cdot \frac{\pi}{2}}\right)^6 = (2\sqrt{5})^6 \cdot e^{i \cdot 3\pi}
   \]
   - Simplify the modulus: \( (2\sqrt{5})^6 = 4000 \).
   - For the argument, \( 3\pi \) is equivalent to \( \pi \) (mod \( 2\pi \)).

   Therefore:
   \[
   (-4 + 2i)^6 = 4000 \cdot e^{i \cdot \pi} = -4000
   \]

### Step 3: Add the Results

Now we add the two results:
\[
(-1 - i)^9 + (-4 + 2i)^6 = (-16 - 16i) + (-4000) = -4016 - 16i
\]

### Final Answer
\[
\boxed{-4016 - 16i}
\]

Would you like further details on any step, or do you have questions?

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Here are five related questions to practice:
1. What is the polar form of \( 1 + i \)?
2. How would you simplify \( (3 + 4i)^4 \)?
3. Can you express \( -2i \) in polar form and raise it to a power?
4. What is the modulus of \( -3 + 3\sqrt{3}i \)?
5. How do you convert a complex number in polar form to rectangular form?

**Tip:** When converting arguments, ensure they fall within the principal range \( [0, 2\pi) \) or \( (-\pi, \pi] \) for consistency.

This page provides a step-by-step solution to the complex expression (-1 - i)^9 + (-4 + 2i)^6. Explore key concepts, including complex numbers in polar form, modulus and argument calculations, and using De Moivre's Theorem to simplify exponentiation.

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