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.