To solve \[ (-1 - i)^9 + (-4 + 2i)^6, \] we can break it into two parts and evaluate each complex number raised to the respective powers.

### Part 1: \( (-1 - i)^9 \)

1. **Convert to polar form**: Let \( z = -1 - i \).
   - The modulus \( |z| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2} \).
   - The argument \( \theta = \tan^{-1} \left( \frac{-1}{-1} \right) = -\frac{3\pi}{4} \).

   So, in polar form: 
   \[
   z = \sqrt{2} \text{cis} \left( -\frac{3\pi}{4} \right),
   \]
   where \(\text{cis}(\theta) = \cos(\theta) + i \sin(\theta)\).

2. **Apply De Moivre's Theorem**:
   \[
   (-1 - i)^9 = \left( \sqrt{2} \right)^9 \text{

Discover how to solve the complex expression (-1 - i)^9 + (-4 + 2i)^6 by converting to polar form and applying De Moivre's Theorem. This problem demonstrates key concepts in complex numbers, exponentiation, and polar form conversion. Ideal for advanced high school or undergraduate students.

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